Gravitation sum related to Centripetal Acceleration

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Homework Help Overview

The problem involves calculating the centripetal acceleration of the moon, given the distance between the Earth and the moon and the acceleration due to gravity on Earth's surface. The context is centered around gravitational forces and centripetal motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal acceleration and gravitational acceleration, with some noting the lack of mass information for the Earth. There are attempts to clarify how to apply gravitational equations at different distances from the Earth.

Discussion Status

Some guidance has been provided regarding the application of gravitational equations and the proportionality of acceleration to distance. Participants are exploring different interpretations of the problem and how to approach the calculations.

Contextual Notes

There is a mention of the assumption that the gravitational field of the moon is negligible for objects on Earth's surface. Additionally, there are reminders about posting guidelines related to mathematical expressions.

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.
IMG_20180826_101415.jpg
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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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