Acceleration of oneself to the moon

1. Sep 21, 2009

psilovethomas

1. The problem statement, all variables and given/known data
F=GmM/r^2

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

3. The attempt at a solution
F=((1.63m/s^2)(109kg)(7.36x10^22kg))/(384,403,000)^3

Am I doing this correctly?

2. Sep 21, 2009

rock.freak667

What are you trying to find exactly?

The force between you and the moon?

3. Sep 21, 2009

Janus

Staff Emeritus
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.

4. Sep 21, 2009

psilovethomas

Yes, or also said the attraction between me and the moon

5. Sep 21, 2009

psilovethomas

Then what is the appropriate G?

6. Sep 21, 2009

psilovethomas

I cannot find a different value for G

7. Sep 21, 2009

rock.freak667

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

8. Sep 21, 2009

psilovethomas

Why is it not the gravitational force of the moon?

9. Sep 21, 2009

rock.freak667

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.

10. Sep 22, 2009

Janus

Staff Emeritus
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

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

11. Sep 22, 2009