Table of contents

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

8. Centripetal Forces & Gravitation

Newton's Law of Gravity

Physics

8. Centripetal Forces & Gravitation

Newton's Law of Gravity

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Multiple Choice

The acceleration due to gravity on the Moon's surface is about one-sixth that on Earth. Given that the radius of the Moon is roughly one-quarter that of the Earth, what is the mass of the Moon in terms of the mass of the Earth?

1/96 of the Earth's mass

1/4 of the Earth's mass

1/24 of the Earth's mass

1/6 of the Earth's mass

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Verified step by step guidance

Start by recalling the formula for gravitational acceleration: \( g = \frac{G \cdot M}{R^2} \), where \( g \) is the acceleration due to gravity, \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.

Given that the acceleration due to gravity on the Moon is one-sixth that on Earth, we can express this as \( g_{\text{Moon}} = \frac{1}{6} g_{\text{Earth}} \).

Also, the radius of the Moon is one-quarter that of the Earth, so \( R_{\text{Moon}} = \frac{1}{4} R_{\text{Earth}} \).

Substitute these relationships into the gravitational acceleration formula for both the Earth and the Moon: \( \frac{G \cdot M_{\text{Moon}}}{(\frac{1}{4} R_{\text{Earth}})^2} = \frac{1}{6} \cdot \frac{G \cdot M_{\text{Earth}}}{R_{\text{Earth}}^2} \).

Simplify the equation to solve for \( M_{\text{Moon}} \) in terms of \( M_{\text{Earth}} \). This involves canceling out common terms and rearranging to find that \( M_{\text{Moon}} = \frac{1}{96} M_{\text{Earth}} \).