Acceleration of oneself to the moon

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Homework Help Overview

The discussion revolves around the gravitational force between an individual and the Moon, utilizing the formula F=GmM/r^2. Participants are examining the correct application of gravitational constants and the distinction between gravitational force and acceleration due to gravity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the original poster's use of the gravitational constant G versus the acceleration due to gravity g. There is an exploration of what exactly is being calculated, whether it is the gravitational force or the attraction between the individual and the Moon.

Discussion Status

There is an ongoing clarification regarding the definitions of G and g, with some participants suggesting that the original poster may need to reconsider the equations being used. Multiple interpretations of the problem are being explored, particularly around the distinction between gravitational force and acceleration.

Contextual Notes

Participants note that the original poster has not provided a value for G, which is leading to confusion in the calculations. There is also mention of the need to differentiate between weight and acceleration in the context of the problem.

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Homework Statement


F=GmM/r^2


Homework Equations


mass of self=109kg
mass of moon=7.36x10^22kg
r=384,403,000m
gravity of moon=1.63m/s^2



The Attempt at a Solution


F=((1.63m/s^2)(109kg)(7.36x10^22kg))/(384,403,000)^3

Am I doing this correctly?
 
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What are you trying to find exactly?

The force between you and the moon?
 
No. The G in the formula you are using is the gravitational constant of the universe, not the acceleration due to gravity on the Moon.
 
rock.freak667 said:
What are you trying to find exactly?

The force between you and the moon?

Yes, or also said the attraction between me and the moon
 
Janus said:
No. The G in the formula you are using is the gravitational constant of the universe, not the acceleration due to gravity on the Moon.

Then what is the appropriate G?
 
I cannot find a different value for G
 
G = universal gravitational constant = 6.67x10-27 m3 kg-1 s-1
 
rock.freak667 said:
G = universal gravitational constant = 6.67x10-27 m3 kg-1 s-1

Why is it not the gravitational force of the moon?
 
psilovethomas said:
Why is it not the gravitational force of the moon?

g is acceleration due to gravity given as F/m (on the moon in your case works out as 1.63m/s2)

But in the formula

[tex]F=G \frac{m_1 m_2}{r^2}[/tex]


G is universal gravitational constant.
 
  • #10
psilovethomas said:
Why is it not the gravitational force of the moon?

The value you are using is the acceleration due to gravity on the Moon, which is a different quanity. It can be found by

[tex]g = \frac{GM}{r^2}[/tex]

You cannot use "g" in the formula you had instead of "G", as, for one reason, the units don't work out.

in your attempt at a solution you have :

[tex]\frac{\frac{m}{s^2} (kg)(kg)}{m^2}[/tex]

which reduces to

[tex]\frac{kg^2}{s^2 m}[/tex]

while the force, which you are trying to find, is measured in

[tex]\frac{kg m}{s^2}[/tex]
 
  • #11
Hello thomas as others have pointed out you are getting G mixed up with g.In the question you are not given the value of G(although it is easy to look this up) so I assume that the intention was for you to use a different equation.In fact if it is the acceleration you want to find, instead of the weight, you do not need an equation at all.Can I suggest that you look up the meaning of g and if you get stuck come back here.
 

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