# 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