# Problem Physics I : gravitation

1. Sep 10, 2008

### fluidistic

1. The problem statement, all variables and given/known data
Suppose that the Earth is a sphere with a radius of $$6371 \text{ km}$$ and that its uniform density is worth $$5517 \text{ kg}/m^3$$. Suppose also that the acceleration of the gravity on its surface is $$g=9.80665 m/s^2$$. Calculate the value of the universal gravitation constant.
(The answer should be $$G=6.672 \cdot 10^{-11}Nm^2/kg^2$$.)

2. Relevant equations $$F_g=\frac{GM_Em}{R_E^2}$$.

3. The attempt at a solution
Using simple very well known formulae, I could determine the mass of the Earth to be about $$5.97 \cdot 10^{24}kg$$.
From $$F_g=\frac{GM_Em}{R_E^2}$$ I got that $$G=\frac{R_E^2F_g}{M_Em}$$. Now the problem is that I got $$G=6.6607246 \cdot 10^{-11}m^3/(kg^2s^2)=6.6607246 \cdot 10^{-11}Nm^2/kg^2$$ as I should but I reached this because I supposed that in the formula m=1kg and the body whose mass is 1kg is on the ground of the Earth. Why do I reach the result when I supposed that there is a mass of 1kg on the ground? Because to use the formula you have to have 2 bodies, the Earth and another body. In my case I supposed it was a body with a mass of 1kg and it worked. But if it had a different mass the result would have been totally different. Also, there's no mention of another body (nor even the formula to calculate the universal gravitational constant) in the statement of the problem. I'm certainly missing plenty of things... Could you explain to me what I don't understand? Thanks in advance.

2. Sep 10, 2008

### alphysicist

Hi fluidistic,

What did you plug in for $F_g$ when you solved for $G$? If you did that correctly I think you'll see why it doesn't matter what $m$ is.

3. Sep 10, 2008

### fluidistic

I plugged $$9.80665m/s^2$$ for $$F_c$$. I know that there is "m" at the denominator but as it is a mass, its unit is not m but kg. So I still don't see why it doesn't matter what m is...

4. Sep 10, 2008

### alphysicist

The value $$9.80665m/s^2$$ is an acceleration so it can't be $F_g$. The force $F_g$ is the gravitational force (weight) that the mass $m$ experiences when it is at a place where the graviational accleration is equal to $g$. So for a mass $m$ at the surface of the earth, what would $F_g$ be?

5. Sep 10, 2008

### fluidistic

$$F_g=mg$$ in that case! Thank you so much, I corrected the units and all work perfectly now.