Gravitation sum related to Centripetal Acceleration

AI Thread Summary
The discussion focuses on calculating the centripetal acceleration of the moon, given its distance from Earth. The gravitational acceleration at the Earth's surface is 10 m/s², and the relationship between gravitational acceleration and distance is highlighted, showing that it decreases with the square of the distance. To find the moon's centripetal acceleration, one can apply the gravitational acceleration formula for two distances: at the Earth's surface and at the moon's distance, simplifying the equations to isolate the unknown. The importance of using proper formatting for mathematical expressions in forum posts is also noted. Understanding these concepts is essential for solving the problem accurately.
Urmi
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Homework Statement


The distance between the centres of the Earth and the moon is 60 times the radius of the earth. Calculate the centripetal acceleration of the moon. Acceleration due to gravity on the Earth's surface is 10m/s.

Homework Equations


Centripetal acceleration= v^2/R
Orbital Velocity=√2gR/√2

The Attempt at a Solution


I found out the centripetal acceleration of Earth but I cannot find that of the moon.
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MathLover69 said:
Firstly, notice that they did not give you the mass of the earth(Me) which is necessary if you want to find the acceleration due to the Earth's G-Field.
a = GME/r^2
However, you don't need the actual mass of the Earth since a is proportional to r^(-2) (a = kr^(-2)). Hope this helped. :)
P.S. This is all assuming that the G-Field of the moon is negligible for an object on Earth's surface.
Umm, I'm sorry...can you just elaborate on the answer, like how to find the a?? I'm confused a bit...
 
The centripetal acceleration of the moon equals the gravitational acceleration of the Earth's gravitational field at distance ##r=R_m=60R_e##.
The gravitational acceleration of Earth's gravitational field at distance ##r=R_e## (that is at the surface of the earth) is given equal to ##10m/s^2##

The gravitational acceleration of the Earth's gravitational field at a specific distance r is given by ##a_r=\frac{GM_e}{r^2}##. Apply this equation first for ##r=R_e ## and then for ##r=R_m=60R_e##. So you ll get two equations. Divide the two equations by parts, the ##GM_e## will be simplified, and you ll get an equation with only one unknown, ##a_{R_m}## that is the gravitational acceleration due to Earth's gravitational field at distance ##r=R_m##.
 
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Delta² said:
The centripetal acceleration of the moon equals the gravitational acceleration of the Earth's gravitational field at distance ##r=R_m=60R_e##.
The gravitational acceleration of Earth's gravitational field at distance ##r=R_e## (that is at the surface of the earth) is given equal to ##10m/s^2##

The gravitational acceleration of the Earth's gravitational field at a specific distance r is given by ##a_r=\frac{GM_e}{r^2}##. Apply this equation first for ##r=R_e ## and then for ##r=R_m=60R_e##. So you ll get two equations. Divide the two equations by parts, the ##GM_e## will be simplified, and you ll get an equation with only one unknown, ##a_{R_m}## that is the gravitational acceleration due to Earth's gravitational field at distance ##r=R_m##.
Thanks a ton! It helped a lot :)
 
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PeterDonis said:
@Urmi, in future, please be advised that you cannot post images of handwritten equations. You need to use the PF LaTeX feature to post math directly in the thread. See here for help:

https://www.physicsforums.com/help/latexhelp/
Okayy, I'm new here so I never knew this...I'll be sure to use this next time :)
 
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