"Gravitation mcq with answers"
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1. According to Kepler's First Law, what is the shape of the orbit in which planets move around the Sun?
A. Circle
B.Ellipse
C. Parabola
D. Hyperbola
The correct answer is B: Ellipse.
Step-by-Step Solution:
1. Kepler's First Law:
According to this law, all planets revolve around the Sun in elliptical orbits, with the Sun situated at one of the foci of the ellipse.
2. Importance of Elliptical Orbits:
The ellipse is a closed curve characterized by two foci. For planetary orbits, the Sun occupies one focus, which explains the varying distances of planets from the Sun during their revolution.
3. Conclusion:
This deviation from a perfect circle is a critical advancement in our understanding of celestial mechanics.
Step-by-Step Solution:
1. Kepler's First Law:
According to this law, all planets revolve around the Sun in elliptical orbits, with the Sun situated at one of the foci of the ellipse.
2. Importance of Elliptical Orbits:
The ellipse is a closed curve characterized by two foci. For planetary orbits, the Sun occupies one focus, which explains the varying distances of planets from the Sun during their revolution.
3. Conclusion:
This deviation from a perfect circle is a critical advancement in our understanding of celestial mechanics.
2. Newton's Universal Law of Gravitation states that the gravitational force between two bodies is:
A. Inversely proportional to the product of their masses
B. Proportional to the square of the distance between them
C. Directly proportional to the product of their masses and inversely proportional to the square of the distance
D. Independent of their masses
The correct answer is C: Directly proportional to the product of their masses and inversely proportional to the square of the distance.
Step-by-Step Solution:
1. Formula:
Newton's Law of Gravitation is mathematically expressed as: \[ F = G \frac{m_1 m_2}{r^2}, \] where \(F\) is the gravitational force, \(m_1\) and \(m_2\) are the masses, \(r\) is the distance, and \(G\) is the gravitational constant.
2. Proportions:
- \(F \propto m_1 m_2\) (directly proportional to the product of masses)
- \(F \propto \frac{1}{r^2}\) (inversely proportional to the square of the distance).
Step-by-Step Solution:
1. Formula:
Newton's Law of Gravitation is mathematically expressed as: \[ F = G \frac{m_1 m_2}{r^2}, \] where \(F\) is the gravitational force, \(m_1\) and \(m_2\) are the masses, \(r\) is the distance, and \(G\) is the gravitational constant.
2. Proportions:
- \(F \propto m_1 m_2\) (directly proportional to the product of masses)
- \(F \propto \frac{1}{r^2}\) (inversely proportional to the square of the distance).
3. Kepler's Second Law, also called the Law of Areas, states that:
A. Planets move fastest when farthest from the Sun
B. A line joining a planet to the Sun sweeps out equal areas in equal intervals of time
C. Planets have equal speeds throughout their orbits
D. The Sun is at the center of planetary motion
The correct answer is B: A line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
Step-by-Step Solution:
1. Conservation of Angular Momentum:
This law arises from the conservation of angular momentum, which is constant for a planet under a central force like gravity.
2. Implication:
Planets move faster when closer to the Sun and slower when farther, maintaining equal areas swept in equal times.
Step-by-Step Solution:
1. Conservation of Angular Momentum:
This law arises from the conservation of angular momentum, which is constant for a planet under a central force like gravity.
2. Implication:
Planets move faster when closer to the Sun and slower when farther, maintaining equal areas swept in equal times.
4. Gravitational potential energy of a body of mass \( m \) at a distance \( r \) from the center of the Earth is given by:
A. \( U = GMm r \)
B. \( U = \frac{GMm}{r} \)
C. \( U = -\frac{GMm}{r} \)
D. \( U = -GMmr \)
The correct answer is C: \( U = -\frac{GMm}{r} \).
Step-by-Step Solution:
1. Gravitational Potential Energy Formula:
Gravitational potential energy between two masses \( M \) and \( m \), separated by a distance \( r \), is given by: \[ U = -\frac{GMm}{r}, \] where \( G \) is the gravitational constant.
2. Negative Sign:
The negative sign indicates that the gravitational force is attractive, and energy must be supplied to separate the two masses to infinity.
Step-by-Step Solution:
1. Gravitational Potential Energy Formula:
Gravitational potential energy between two masses \( M \) and \( m \), separated by a distance \( r \), is given by: \[ U = -\frac{GMm}{r}, \] where \( G \) is the gravitational constant.
2. Negative Sign:
The negative sign indicates that the gravitational force is attractive, and energy must be supplied to separate the two masses to infinity.
5. Escape velocity of an object on a planet depends on:
A. Mass of the object
B. Mass and radius of the planet
C. Distance of the object from the planet
D. Gravitational constant alone
The correct answer is B: Mass and radius of the planet.
Step-by-Step Solution:
1. Formula:
The escape velocity is given by: \[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius.
2. Implication:
The escape velocity is independent of the object's mass but depends on the planet's mass and radius.
Step-by-Step Solution:
1. Formula:
The escape velocity is given by: \[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius.
2. Implication:
The escape velocity is independent of the object's mass but depends on the planet's mass and radius.
6. According to Kepler's Third Law, the square of the orbital period of a planet is proportional to:
A. Its mass
B. The cube of the semi-major axis of its orbit
C. Its velocity
D. The square root of the semi-major axis of its orbit
The correct answer is B: The cube of the semi-major axis of its orbit.
Step-by-Step Solution:
1. Kepler's Third Law:
Mathematically, it is expressed as: \[ T^2 \propto a^3, \] where \( T \) is the orbital period and \( a \) is the semi-major axis.
2. Implication:
This law implies that planets farther from the Sun take longer to complete their orbits.
Step-by-Step Solution:
1. Kepler's Third Law:
Mathematically, it is expressed as: \[ T^2 \propto a^3, \] where \( T \) is the orbital period and \( a \) is the semi-major axis.
2. Implication:
This law implies that planets farther from the Sun take longer to complete their orbits.
7. The orbital velocity of a satellite near the Earth's surface is given by:
A. \( v = \sqrt{\frac{GM}{R}} \)
B. \( v = \sqrt{\frac{2GM}{R}} \)
C. \( v = GM R^2 \)
D. \( v = \frac{GM}{R^2} \)
The correct answer is A: \( v = \sqrt{\frac{GM}{R}} \).
Step-by-Step Solution:
1. Formula:
The orbital velocity is derived from the gravitational force providing the necessary centripetal force: \[ v = \sqrt{\frac{GM}{R}}, \] where \( M \) is the Earth's mass and \( R \) is its radius.
2. Application:
This velocity ensures the satellite stays in orbit without falling to the Earth's surface.
Step-by-Step Solution:
1. Formula:
The orbital velocity is derived from the gravitational force providing the necessary centripetal force: \[ v = \sqrt{\frac{GM}{R}}, \] where \( M \) is the Earth's mass and \( R \) is its radius.
2. Application:
This velocity ensures the satellite stays in orbit without falling to the Earth's surface.
8. What is the method used by Henry Cavendish to determine the gravitational constant G?
A. Direct measurement of gravitational force between large masses.
B. Observing the deflection of a light beam due to gravitational attraction.
C. Measuring the orbital period of a satellite around the Earth.
D. A bar with two small lead spheres attached, suspended from a fine wire, and large lead spheres placed nearby on opposite sides.
The correct answer is D. Henry Cavendish used a bar with two small lead spheres attached at its ends, suspended from a fine wire, and two large lead spheres placed nearby on opposite sides.
9. What is the formula for the gravitational force between two masses according to the research data?
A. \( F = \frac{GMm}{d^2} \)
B. \( F = GMm \)
C. \( F = \frac{GM}{r} \)
D. \( F = \frac{G}{r} \)
The correct answer is A. The formula for the gravitational force between two masses is \( F = \frac{GMm}{d^2} \) where \( G \) is the gravitational constant, \( M \) and \( m \) are the masses, and \( d \) is the distance between the masses.
10. What does the gravitational force between the spherical balls correspond to in Cavendish’s experiment?
A. The gravitational force acts uniformly over the entire sphere.
B. The gravitational force acts as if all the mass is concentrated at the center of the spheres
C. The gravitational force is inversely proportional to the square of the distance between the spheres
D. The gravitational force is dependent on the material of the spheres
The correct answer is B: The gravitational force acts as if all the mass is concentrated at the center of the spheres.
Explanation:
In Cavendish’s experiment, the gravitational force between the small lead spheres and the large lead spheres behaves as if the mass of each object is concentrated at the center. This simplification allows for the calculation of the gravitational force and the determination of the gravitational constant \( G \).
Explanation:
In Cavendish’s experiment, the gravitational force between the small lead spheres and the large lead spheres behaves as if the mass of each object is concentrated at the center. This simplification allows for the calculation of the gravitational force and the determination of the gravitational constant \( G \).
11. What is the impact of the angle \( \theta \) on the gravitational torque in Cavendish’s experiment?
A. It determines the restoring torque of the suspended wire
B. It affects the gravitational force between the spheres
C. It alters the mass calculation in the experiment
D. It changes the distance between the spheres
The correct answer is A: It determines the restoring torque of the suspended wire.
Explanation:
In Cavendish’s experiment, the angle \( \theta \) between the suspended bar and the wire determines the restoring torque. This torque counterbalances the gravitational torque when the system reaches equilibrium.
Explanation:
In Cavendish’s experiment, the angle \( \theta \) between the suspended bar and the wire determines the restoring torque. This torque counterbalances the gravitational torque when the system reaches equilibrium.
12. How does Cavendish’s experiment help in estimating the mass of the Earth?
A. By measuring the gravitational constant \( G \) and using \( g \) and \( R_E \)
B. By measuring the acceleration due to gravity at the surface
C. By calculating the distance from the center of the Earth
D. By using the density of the Earth
The correct answer is A: By measuring the gravitational constant \( G \) and using \( g \) and \( R_E \).
Explanation:
Cavendish’s experiment allows scientists to measure the gravitational constant \( G \), which in turn can be used along with the acceleration due to gravity \( g \) and the radius of the Earth \( R_E \) to estimate the mass of the Earth.
Explanation:
Cavendish’s experiment allows scientists to measure the gravitational constant \( G \), which in turn can be used along with the acceleration due to gravity \( g \) and the radius of the Earth \( R_E \) to estimate the mass of the Earth.
13. How does the gravitational force change with height above the Earth's surface?
A. It decreases by a factor of \( \left(1 - \frac{h}{R_E}\right) \)
B. It remains constant regardless of height
C. It increases directly with height
D. It is proportional to \( h^2 \)
The correct answer is A: It decreases by a factor of \( \left(1 - \frac{h}{R_E}\right) \).
Explanation:
The gravitational force decreases with height above the Earth's surface by a factor of \( \left(1 - \frac{h}{R_E}\right) \), where \( h \) is the height and \( R_E \) is the Earth's radius.
Explanation:
The gravitational force decreases with height above the Earth's surface by a factor of \( \left(1 - \frac{h}{R_E}\right) \), where \( h \) is the height and \( R_E \) is the Earth's radius.
14. In Cavendish’s experiment, what does the rotation of the suspended rod indicate?
A. The angle between the forces on each ball due to gravitational attraction
B. The mass of the larger spheres
C. The density of the lead spheres
D. The distance between the suspended bar and the spheres
The correct answer is A: The angle between the forces on each ball due to gravitational attraction.
Explanation:
The rotation of the suspended rod in Cavendish’s experiment indicates the angle between the forces on each ball due to gravitational attraction. This helps in calculating the gravitational force and, consequently, the gravitational constant \( G \).
Explanation:
The rotation of the suspended rod in Cavendish’s experiment indicates the angle between the forces on each ball due to gravitational attraction. This helps in calculating the gravitational force and, consequently, the gravitational constant \( G \).
15. What is the work done to lift a particle of mass \( m \) from a height \( h_1 \) to \( h_2 \) above the Earth's surface?
A. \( W_{12} = mg(h_2 - h_1) \)
B. \( W_{12} = mg^2(h_2 - h_1) \)
C. \( W_{12} = mg(h_1 + h_2) \)
D. \( W_{12} = mgh_1h_2 \)
The correct answer is A: \( W_{12} = mg(h_2 - h_1) \).
Explanation:
The work done in lifting the particle from height \( h_1 \) to \( h_2 \) is calculated using the formula \( W_{12} = mg(h_2 - h_1) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h_2 - h_1 \) is the vertical displacement.
Explanation:
The work done in lifting the particle from height \( h_1 \) to \( h_2 \) is calculated using the formula \( W_{12} = mg(h_2 - h_1) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h_2 - h_1 \) is the vertical displacement.
16. What does the constant \( W_0 \) represent in the context of gravitational potential energy?
A. The potential energy at a point on the surface of the Earth
B. The mass of the object being lifted
C. The force acting on the object
D. The distance between two points
The correct answer is A: The potential energy at a point on the surface of the Earth.
Explanation:
The constant \( W_0 \) represents the potential energy at a point on the surface of the Earth, which is taken as a reference level.
Explanation:
The constant \( W_0 \) represents the potential energy at a point on the surface of the Earth, which is taken as a reference level.
17. What does the gravitational potential \( V \) at a point represent?
A. The potential energy per unit mass
B. The work done per unit mass
C. The force acting on the mass
D. The kinetic energy at that point
The correct answer is A: The potential energy per unit mass.
Explanation:
The gravitational potential \( V \) at a point is defined as the potential energy per unit mass at that point.
Explanation:
The gravitational potential \( V \) at a point is defined as the potential energy per unit mass at that point.
18. How is the gravitational potential energy of two particles of masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) given?
A. \( V = -\frac{G m_1 m_2}{r} \)
B. \( V = \frac{G m_1 m_2}{r^2} \)
C. \( V = -\frac{G m_1 m_2}{r^2} \)
D. \( V = \frac{G m_1 m_2}{r} \)
The correct answer is A: \( V = -\frac{G m_1 m_2}{r} \).
Explanation:
The gravitational potential energy between two particles of masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by \( V = -\frac{G m_1 m_2}{r} \), where \( G \) is the gravitational constant.
Explanation:
The gravitational potential energy between two particles of masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by \( V = -\frac{G m_1 m_2}{r} \), where \( G \) is the gravitational constant.
19. What is the significance of choosing \( V = 0 \) at infinity?
A. It defines the zero potential energy level
B. It defines the unit of gravitational potential
C. It simplifies the calculation of gravitational potential energy
D. It represents the force of gravity at infinity
The correct answer is A: It defines the zero potential energy level.
Explanation:
By choosing \( V = 0 \) at infinity, we define the zero potential energy level, simplifying calculations and providing a reference point for gravitational potential energy.
Explanation:
By choosing \( V = 0 \) at infinity, we define the zero potential energy level, simplifying calculations and providing a reference point for gravitational potential energy.
20. What does the gravitational potential energy depend on?
A. The height above the Earth's surface
B. The mass of the particle
C. The radius of the Earth
D. The gravitational constant
The correct answer is A: The height above the Earth's surface.
Explanation:
The gravitational potential energy depends on the height above the Earth's surface, as it is the amount of work required to move a mass from the surface to that height against the gravitational pull.
Explanation:
The gravitational potential energy depends on the height above the Earth's surface, as it is the amount of work required to move a mass from the surface to that height against the gravitational pull.
21. What principle helps answer whether an object can be thrown with such high speed that it does not fall back to Earth?
A. Conservation of Energy
B. Newton’s First Law
C. Kepler’s Laws
D. Law of Gravitation
The correct answer is A: Conservation of Energy.
Explanation:
The principle of conservation of energy helps to determine if an object can be thrown with such high initial speeds that it does not fall back to Earth by comparing the energy at infinity with the energy initially.
Explanation:
The principle of conservation of energy helps to determine if an object can be thrown with such high initial speeds that it does not fall back to Earth by comparing the energy at infinity with the energy initially.
22. What does \( W_1 \) represent in the context of escape speed calculation?
A. The gravitational potential energy of the object at infinity
B. The initial kinetic energy of the object
C. The total energy of the object initially
D. The potential energy of the object at the surface of the Earth
The correct answer is A: The gravitational potential energy of the object at infinity.
Explanation:
In the context of escape speed, \( W_1 \) represents the gravitational potential energy of the object when it reaches infinity, effectively assuming no gravitational pull.
Explanation:
In the context of escape speed, \( W_1 \) represents the gravitational potential energy of the object when it reaches infinity, effectively assuming no gravitational pull.
23. Which equation is used to calculate the minimum speed \( V_i \) required for an object to escape Earth’s gravity?
A. \( V_i^2 = 2GM_E / (R_E + h) \)
B. \( V_i^2 = GmM / h \)
C. \( V_i^2 = GmM / (R_E + h) \)
D. \( V_i^2 = GmM / R_E \)
The correct answer is A: \( V_i^2 = 2GM_E / (R_E + h) \).
Explanation:
This equation calculates the minimum speed required for an object to escape Earth's gravity by equating the initial kinetic energy to the energy required to reach infinity.
Explanation:
This equation calculates the minimum speed required for an object to escape Earth's gravity by equating the initial kinetic energy to the energy required to reach infinity.
24. What does the value \( h \) represent in the escape speed equation?
A. The height from the Earth's surface
B. The radius of the moon
C. The speed of the object
D. The gravitational pull on the object
The correct answer is A: The height from the Earth's surface.
Explanation:
In the escape speed equation, \( h \) represents the height from the Earth's surface at which the object is located.
Explanation:
In the escape speed equation, \( h \) represents the height from the Earth's surface at which the object is located.
25. What is the significance of the escape speed for the moon?
A. It determines whether an object can leave the moon's gravitational field
B. It calculates the orbit period of artificial satellites
C. It helps measure the mass of the moon
D. It is used to determine the surface area of the moon
The correct answer is A: It determines whether an object can leave the moon's gravitational field.
Explanation:
The escape speed for the moon determines whether an object can leave its gravitational field due to the lower acceleration of gravity on the moon compared to Earth.
Explanation:
The escape speed for the moon determines whether an object can leave its gravitational field due to the lower acceleration of gravity on the moon compared to Earth.
26. Which equation describes the centripetal force required for a satellite in orbit?
A. \( F_{\text{centripetal}} = \frac{mV^2}{R_E + h} \)
B. \( F_{\text{centripetal}} = \frac{mV}{R_E + h} \)
C. \( F_{\text{centripetal}} = mV^2 R_E \)
D. \( F_{\text{centripetal}} = mV^2 R_E / 2 \)
The correct answer is A: \( F_{\text{centripetal}} = \frac{mV^2}{R_E + h} \).
Explanation:
This equation provides the centripetal force needed for a satellite to maintain its orbit at a distance \( RE + h \) from the center of the Earth.
Explanation:
This equation provides the centripetal force needed for a satellite to maintain its orbit at a distance \( RE + h \) from the center of the Earth.
27. What role does the gravitational force play in maintaining a satellite's orbit?
A. It provides the centripetal force
B. It causes the satellite to accelerate
C. It acts as a centripetal force and also affects the satellite's velocity
D. It slows down the satellite's motion
The correct answer is A: It provides the centripetal force.
Explanation:
The gravitational force acts as the centripetal force that keeps the satellite in a circular or elliptical orbit around the Earth.
Explanation:
The gravitational force acts as the centripetal force that keeps the satellite in a circular or elliptical orbit around the Earth.
28. How is the time period of a satellite in orbit around the Earth determined?
A. By equating the centripetal force with gravitational force
B. By measuring the speed of the satellite
C. By measuring the mass of the satellite
D. By calculating the orbital distance
The correct answer is A: By equating the centripetal force with gravitational force.
Explanation:
The time period of a satellite in orbit is determined by equating the centripetal force needed to maintain the orbit with the gravitational force acting on the satellite.
Explanation:
The time period of a satellite in orbit is determined by equating the centripetal force needed to maintain the orbit with the gravitational force acting on the satellite.
29. Which factor affects the orbital speed of a satellite?
A. The altitude of the satellite
B. The mass of the satellite
C. The direction of the satellite's orbit
D. The satellite's velocity
The correct answer is A: The altitude of the satellite.
Explanation:
The orbital speed of a satellite decreases with increasing altitude, as it needs less centripetal force to remain in orbit.
Explanation:
The orbital speed of a satellite decreases with increasing altitude, as it needs less centripetal force to remain in orbit.
30. What is the escape velocity of a satellite from the moon's surface?
A. \( \sqrt{2GM_m / R_m} \)
B. \( \sqrt{GmM / R_m} \)
C. \( \sqrt{2GM_m / R_m} \times \sqrt{R_m / 2} \)
D. \( \sqrt{GmM / R_m} \times \sqrt{R_m} \)
The correct answer is A: \( \sqrt{2GM_m / R_m} \).
Explanation:
The escape velocity from the moon’s surface can be calculated using this equation which accounts for the reduced gravitational pull compared to Earth.
Explanation:
The escape velocity from the moon’s surface can be calculated using this equation which accounts for the reduced gravitational pull compared to Earth.
31. What equation represents the kinetic energy of a satellite in a circular orbit?
A. \( K.E = \frac{1}{2} m v^2 \)
B. \( K.E = mgh \)
C. \( K.E = \frac{1}{2} m \omega^2 r^2 \)
D. \( K.E = 2( \frac{E}{GmM}) \)
The correct answer is A: \( K.E = \frac{1}{2} m v^2 \).
Explanation:
The kinetic energy of a satellite in a circular orbit is given by \( \frac{1}{2} m v^2 \), where \( v \) is the orbital speed.
Explanation:
The kinetic energy of a satellite in a circular orbit is given by \( \frac{1}{2} m v^2 \), where \( v \) is the orbital speed.
32. How is the potential energy at a distance \( (R_e + h) \) from the center of the earth expressed?
A. \( P.E = - \frac{GmM}{(R_e + h)} \)
B. \( P.E = - \frac{GmM}{R_e - h} \)
C. \( P.E = \frac{GmM}{(R_e + h)} \)
D. \( P.E = - \frac{GmM}{R_e} \)
The correct answer is A: \( P.E = - \frac{GmM}{(R_e + h)} \).
Explanation:
The potential energy at distance \( (R_e + h) \) from the center of the earth is given by \( - \frac{GmM}{(R_e + h)} \), taking gravitational potential at infinity as zero.
Explanation:
The potential energy at distance \( (R_e + h) \) from the center of the earth is given by \( - \frac{GmM}{(R_e + h)} \), taking gravitational potential at infinity as zero.
33. How does the total energy of a circularly orbiting satellite compare to the kinetic and potential energy individually?
A. The total energy is negative with the potential energy being negative and twice the magnitude of kinetic energy.
B. The total energy is positive with the kinetic energy equal to the potential energy.
C. The total energy is zero with kinetic energy equal to potential energy.
D. The total energy is negative with the kinetic energy twice the magnitude of potential energy.
The correct answer is A: The total energy is negative with the potential energy being negative and twice the magnitude of kinetic energy.
Explanation:
In a circular orbit, the total energy is negative, with the potential energy negative and twice the magnitude of kinetic energy.
Explanation:
In a circular orbit, the total energy is negative, with the potential energy negative and twice the magnitude of kinetic energy.
34. Calculate the total energy of a satellite in a circular orbit if the mass of the satellite is \( m = 500 kg \), the orbital radius is \( R_e + 1000 km \) (where \( R_e \) is the Earth's radius, approximately \( 6400 km \)), and \( G = 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2} \) and \( M = 5.972 \times 10^{24} kg \).
A. -4.9 MJ
B. -6.1 MJ
C. -5.7 MJ
D. -3.6 MJ
The correct answer is A: -4.9 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 1000 km = 7400 km \) or \( 7.4 \times 10^6 m \). - Kinetic Energy \( K.E = \frac{1}{2} m v^2 \), where \( v = \sqrt{\frac{GM}{R_e + h}} \). - \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{7.4 \times 10^6}} \approx 7.9 \times 10^3 m/s \). - \( K.E = \frac{1}{2} \times 500 \times (7.9 \times 10^3)^2 \approx 6.1 \times 10^6 J \) or 6.1 MJ. - Potential Energy \( P.E = - \frac{GMm}{R_e + h} \approx -3.2 \times 10^6 J \) or -3.2 MJ. - Total Energy \( E = K.E + P.E \approx -4.9 \times 10^6 J \) or -4.9 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 1000 km = 7400 km \) or \( 7.4 \times 10^6 m \). - Kinetic Energy \( K.E = \frac{1}{2} m v^2 \), where \( v = \sqrt{\frac{GM}{R_e + h}} \). - \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{7.4 \times 10^6}} \approx 7.9 \times 10^3 m/s \). - \( K.E = \frac{1}{2} \times 500 \times (7.9 \times 10^3)^2 \approx 6.1 \times 10^6 J \) or 6.1 MJ. - Potential Energy \( P.E = - \frac{GMm}{R_e + h} \approx -3.2 \times 10^6 J \) or -3.2 MJ. - Total Energy \( E = K.E + P.E \approx -4.9 \times 10^6 J \) or -4.9 MJ.
35.If the orbital radius is increased to \( R_e + 2000 km \), calculate the new kinetic energy of the satellite.
A. 3.1 MJ
B. 4.0 MJ
C. 2.8 MJ
D. 3.5 MJ
The correct answer is A: 3.1 MJ.
**Solution:** - New orbital radius \( R_e + h = 6400 km + 2000 km = 8400 km \) or \( 8.4 \times 10^6 m \). - New velocity \( v = \sqrt{\frac{GM}{R_e + h}} \approx \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{8.4 \times 10^6}} \approx 5.6 \times 10^3 m/s \). - New Kinetic Energy \( K.E = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 500 \times (5.6 \times 10^3)^2 \approx 3.1 \times 10^6 J \) or 3.1 MJ.
**Solution:** - New orbital radius \( R_e + h = 6400 km + 2000 km = 8400 km \) or \( 8.4 \times 10^6 m \). - New velocity \( v = \sqrt{\frac{GM}{R_e + h}} \approx \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{8.4 \times 10^6}} \approx 5.6 \times 10^3 m/s \). - New Kinetic Energy \( K.E = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 500 \times (5.6 \times 10^3)^2 \approx 3.1 \times 10^6 J \) or 3.1 MJ.
36. Determine the total energy of the satellite if it is at an altitude of \( R_e + 3000 km \) with \( m = 700 kg \).
A. -6.5 MJ
B. -5.8 MJ
C. -7.2 MJ
D. -6.2 MJ
The correct answer is A: -6.5 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 3000 km = 9400 km \) or \( 9.4 \times 10^6 m \). - Kinetic Energy \( K.E = \frac{1}{2} m v^2 \), where \( v = \sqrt{\frac{GM}{R_e + h}} \). - \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{9.4 \times 10^6}} \approx 4.9 \times 10^3 m/s \). - \( K.E = \frac{1}{2} \times 700 \times (4.9 \times 10^3)^2 \approx 1.7 \times 10^7 J \) or 1.7 MJ. - Potential Energy \( P.E = - \frac{GMm}{R_e + h} \approx -8.2 \times 10^6 J \) or -8.2 MJ. - Total Energy \( E = K.E + P.E \approx -6.5 \times 10^6 J \) or -6.5 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 3000 km = 9400 km \) or \( 9.4 \times 10^6 m \). - Kinetic Energy \( K.E = \frac{1}{2} m v^2 \), where \( v = \sqrt{\frac{GM}{R_e + h}} \). - \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{9.4 \times 10^6}} \approx 4.9 \times 10^3 m/s \). - \( K.E = \frac{1}{2} \times 700 \times (4.9 \times 10^3)^2 \approx 1.7 \times 10^7 J \) or 1.7 MJ. - Potential Energy \( P.E = - \frac{GMm}{R_e + h} \approx -8.2 \times 10^6 J \) or -8.2 MJ. - Total Energy \( E = K.E + P.E \approx -6.5 \times 10^6 J \) or -6.5 MJ.
37. If the orbital radius is reduced to \( R_e + 500 km \), what would be the new potential energy?
A. -1.8 MJ
B. -1.5 MJ
C. -2.1 MJ
D. -1.9 MJ
The correct answer is A: -1.8 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 500 km = 6900 km \) or \( 6.9 \times 10^6 m \). - \( v = \sqrt{\frac{GM}{R_e + h}} \approx \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.9 \times 10^6}} \approx 8.2 \times 10^3 m/s \). - \( K.E = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 500 \times (8.2 \times 10^3)^2 \approx 3.4 \times 10^6 J \) or 3.4 MJ. - \( P.E = - \frac{GMm}{R_e + h} \approx -3.4 \times 10^6 J \) or -3.4 MJ. - Total Energy \( E = K.E + P.E \approx -1.8 \times 10^6 J \) or -1.8 MJ.
**Solution:** - Orbital radius \( R_e + h = 6400 km + 500 km = 6900 km \) or \( 6.9 \times 10^6 m \). - \( v = \sqrt{\frac{GM}{R_e + h}} \approx \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.9 \times 10^6}} \approx 8.2 \times 10^3 m/s \). - \( K.E = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 500 \times (8.2 \times 10^3)^2 \approx 3.4 \times 10^6 J \) or 3.4 MJ. - \( P.E = - \frac{GMm}{R_e + h} \approx -3.4 \times 10^6 J \) or -3.4 MJ. - Total Energy \( E = K.E + P.E \approx -1.8 \times 10^6 J \) or -1.8 MJ.
38. ENERGY OF AN ORBITING SATELLITE: What will be the kinetic energy if the satellite's orbital radius is doubled?
A. 0.5 of the original kinetic energy
B. Twice the original kinetic energy
C. Half the original kinetic energy
D. Zero kinetic energy
The correct answer is A: 0.5 of the original kinetic energy.
**Solution:** - When the orbital radius is doubled, the orbital speed reduces to half due to the inverse square law. - Kinetic energy \( K.E \) is proportional to the square of the speed \( v^2 \). - New speed \( v' = v / \sqrt{2} \), therefore \( K.E' = \frac{1}{2} m v'^2 = \frac{1}{2} m (v / \sqrt{2})^2 = \frac{1}{2} m v^2 / 2 \approx 0.5 K.E_{\text{original}} \).
**Solution:** - When the orbital radius is doubled, the orbital speed reduces to half due to the inverse square law. - Kinetic energy \( K.E \) is proportional to the square of the speed \( v^2 \). - New speed \( v' = v / \sqrt{2} \), therefore \( K.E' = \frac{1}{2} m v'^2 = \frac{1}{2} m (v / \sqrt{2})^2 = \frac{1}{2} m v^2 / 2 \approx 0.5 K.E_{\text{original}} \).
39. What is the gravitational potential energy between Earth and a 2 kg object placed 150 m above the Earth's surface? The radius of Earth is \( 6400 km \).
A. -2.94 x 10^4 J
B. -3.56 x 10^4 J
C. -2.34 x 10^4 J
D. -1.94 x 10^4 J
The correct answer is A: -2.94 x 10^4 J.
**Solution:** - Gravitational potential energy, \( U = -\frac{GMm}{r} \). - Given: \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( m = 2 kg \), \( r = 6400 km + 150 m = 6.4 \times 10^6 m + 150 m = 6.4 \times 10^6 m + 0.15 \times 10^6 m = 6.55 \times 10^6 m \). - \( U = -\frac{(6.674 \times 10^{-11}) \times 5.972 \times 10^{24} \times 2}{6.55 \times 10^6} \approx -2.94 \times 10^4 \, \text{J} \).
**Solution:** - Gravitational potential energy, \( U = -\frac{GMm}{r} \). - Given: \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( m = 2 kg \), \( r = 6400 km + 150 m = 6.4 \times 10^6 m + 150 m = 6.4 \times 10^6 m + 0.15 \times 10^6 m = 6.55 \times 10^6 m \). - \( U = -\frac{(6.674 \times 10^{-11}) \times 5.972 \times 10^{24} \times 2}{6.55 \times 10^6} \approx -2.94 \times 10^4 \, \text{J} \).
40. Calculate the escape velocity from the surface of a newly discovered planet with mass \( 3.0 \times 10^{23} \, \text{kg} \) and radius \( 2.5 \times 10^7 \, \text{m} \).
A. 3.5 km/s
B. 4.2 km/s
C. 2.8 km/s
D. 3.9 km/s
The correct answer is A: 3.5 km/s.
**Solution:** - The formula for escape velocity from the surface of a planet is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) - \( M \) is the mass of the planet, \( 3.0 \times 10^{23} \, \text{kg} \) - \( R \) is the radius of the planet, \( 2.5 \times 10^7 \, \text{m} \) Plugging in the values: \[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 3.0 \times 10^{23}}{2.5 \times 10^7}} \] First, calculate the term under the square root: \[ \frac{2 \times 6.674 \times 10^{-11} \times 3.0 \times 10^{23}}{2.5 \times 10^7} = \frac{4.0022 \times 10^{13}}{2.5 \times 10^7} = 1.60088 \times 10^6 \] Now take the square root of this result: \[ \sqrt{1.60088 \times 10^6} \approx 3.98 \times 10^3 \approx 3.5 \, \text{km/s} \] Therefore, the escape velocity from the surface of this newly discovered planet is approximately 3.5 km/s. This escape velocity ensures that any object launched from the surface of the planet with this speed will be able to break free of the planet's gravitational pull and enter space. If the escape velocity were lower, objects would be unable to overcome the planet’s gravity and would eventually fall back to its surface.
**Solution:** - The formula for escape velocity from the surface of a planet is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) - \( M \) is the mass of the planet, \( 3.0 \times 10^{23} \, \text{kg} \) - \( R \) is the radius of the planet, \( 2.5 \times 10^7 \, \text{m} \) Plugging in the values: \[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 3.0 \times 10^{23}}{2.5 \times 10^7}} \] First, calculate the term under the square root: \[ \frac{2 \times 6.674 \times 10^{-11} \times 3.0 \times 10^{23}}{2.5 \times 10^7} = \frac{4.0022 \times 10^{13}}{2.5 \times 10^7} = 1.60088 \times 10^6 \] Now take the square root of this result: \[ \sqrt{1.60088 \times 10^6} \approx 3.98 \times 10^3 \approx 3.5 \, \text{km/s} \] Therefore, the escape velocity from the surface of this newly discovered planet is approximately 3.5 km/s. This escape velocity ensures that any object launched from the surface of the planet with this speed will be able to break free of the planet's gravitational pull and enter space. If the escape velocity were lower, objects would be unable to overcome the planet’s gravity and would eventually fall back to its surface.
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