Acceleration of oneself to the moon

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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.
 
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]
 
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.