Acceleration of oneself to the moon

[psilovethomas]

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?

 

[rock.freak667]

What are you trying to find exactly?

The force between you and the moon?

 

[Janus]

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.

 

[psilovethomas]

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

 

[psilovethomas]

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?

 

[psilovethomas]

I cannot find a different value for G

 

[rock.freak667]

G = universal gravitational constant = 6.67x10^-27 m³ kg^-1 s^-1

 

[psilovethomas]

G = universal gravitational constant = 6.67x10^-27 m³ kg^-1 s^-1

Why is it not the gravitational force of the moon?

 

[rock.freak667]

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

F=G \frac{m_1 m_2}{r^2}


G is universal gravitational constant.

 

[Janus]

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

g = \frac{GM}{r^2}

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 :

\frac{\frac{m}{s^2} (kg)(kg)}{m^2}

which reduces to

\frac{kg^2}{s^2 m}

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

\frac{kg m}{s^2}

 

[Dadface]

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.