Answer: A) \( 9.8 \, \text{m/s}^2 \), \( 4.099 \times 10^{-25} \, \text{m/s}^2 \)
Solution:
The gravitational force between two masses is given by:
\[
F = G \frac{m_1 m_2}{r^2}
\]
Here, the force acting on the apple due to Earth is:
\[
F = G \frac{M_{\text{Earth}} m_{\text{apple}}}{R_{\text{Earth}}^2}
\]
The acceleration of the apple towards the Earth, \( a_{\text{apple}} \), is:
\[
a_{\text{apple}} = \frac{F}{m_{\text{apple}}} = G \frac{M_{\text{Earth}}}{R_{\text{Earth}}^2}
\]
Substituting the given values:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
- \( M_{\text{Earth}} = 5.983 \times 10^{24} \, \text{kg} \)
- \( R_{\text{Earth}} = 6.378 \times 10^6 \, \text{m} \)
\[
a_{\text{apple}} = 6.67 \times 10^{-11} \cdot \frac{5.983 \times 10^{24}}{(6.378 \times 10^6)^2}
\]
\[
a_{\text{apple}} = 9.8 \, \text{m/s}^2
\]
Now, the acceleration of the Earth towards the apple, \( a_{\text{Earth}} \), is:
\[
a_{\text{Earth}} = \frac{F}{M_{\text{Earth}}}
\]
Using \( F = G \frac{M_{\text{Earth}} m_{\text{apple}}}{R_{\text{Earth}}^2} \):
\[
a_{\text{Earth}} = G \frac{m_{\text{apple}}}{R_{\text{Earth}}^2}
\]
Substituting the values:
\[
a_{\text{Earth}} = 6.67 \times 10^{-11} \cdot \frac{0.25}{(6.378 \times 10^6)^2}
\]
\[
a_{\text{Earth}} = 4.099 \times 10^{-25} \, \text{m/s}^2
\]
Final Answer:
- Acceleration of the apple towards the Earth: \( 9.8 \, \text{m/s}^2 \)
- Acceleration of the Earth towards the apple: \( 4.099 \times 10^{-25} \, \text{m/s}^2 \)
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