Language

Loading…

Physics Playground

Built on open, peer-reviewed sources: OpenStax, The Feynman Lectures on Physics, HyperPhysics, NASA and NIST. See our sources.

Learn

Physics course
Math for Physics
Start with the math toolkit
Simulations
Glossary
Exercises
Past exams
Library
How we teach

Company

Pricing
Terms
Privacy
Contact

Earth’s g at the Moon · Physics Playground

Exercises Circular motion and gravity

Example from Physics, 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity

Earth’s g at the Moon

Find the acceleration due to Earth’s gravity at the distance of the moon.

\begin{array}{l} G = 6 .67 \times 10^{–11} {N · m}^{2} {/kg}^{2} \\ Earth-moon distance = 3.84 \times 10^{8} m \\ Earth’s mass = 5.98 \times 10^{24} kg \end{array}

7.5

Express the force of gravity in terms of g.

F = W = m a = m g

7.6

Combine with the equation for universal gravitation.

m g = m G \frac{M}{r^{2}}

7.7

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Cancel m and substitute.

$g = G \frac{M}{r^{2}} = (6.67 \times 10^{- 11} \frac{N \cdot m^{2}}{{kg}^{2}}) (\frac{5.98 \times 10^{24} kg}{{(3.84 \times 10^{8} m)}^{2}}) = 2.70 \times 10^{- 3} {m/s}^{2}$
7.8

How did it go?

PreviousChange in g Next How Fast Does a Car Tire Spin?

From Physics (1st edition) by OpenStax (Rice University), section 7.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.

Solution