# Gravitational force

1. Aug 4, 2009

### dance_sg

1. The problem statement, all variables and given/known data

When the distance separating two masses, M and m, is 1.2 x 10^10 m, the gravitational force of attraction is 5.0 N. If the mass of M becomes 3M and the separation distance becomes
2.4 x 1010 m, what will be the force?

2. Relevant equations
Fg=MaMb/r^2

3. The attempt at a solution
since i have r and Fg (first part of question) i tried solving for the masses. but how do i solve for both when i need at least one to solve the other/??

2. Aug 4, 2009

### Pengwuino

You're missing the gravitational constant in your equation. I suggest setting up a ratio between the two forces and see what happens.

3. Aug 4, 2009

### dance_sg

huh? im still kinda confused..

4. Aug 4, 2009

### Pengwuino

You know : $$F_1 = \frac{{GM_1 m_1 }}{{r_1 ^2 }}$$ and that $$F_2 = \frac{{GM_2 m_2 }}{{r_2 ^2 }}$$.

What do you get when you divide the second equation by the first, that is: $$\frac{{F_2 }}{{F_1 }}$$.

You know:

$$$\begin{array}{l} M_2 = 3M_1 \\ m_1 = m_2 \\ \end{array}$$$

along with the actual initial force, $$F_1$$. You can solve for the ratio of forces which will tell you what $$F_2$$ is.

5. Aug 5, 2009

### prob_solv

or, use this EASIER way :

for the first condition, and the second condition, the gravitational constant is always the same.
$$G=G$$

You have known what the value of F is :
$$F=\frac{GMm}{r^2 }$$
and, the value of G is :
$$G=\frac{Fr^2 }{Mm}$$

you can make it like this :
$$G=G$$
$$\frac{Fr1^2 }{M1m1}=\frac{Fr2^2 }{M2m2}$$
substitute the variable ( like M2 to 3 Mi ) and you can eliminate the variable. At last, there will be an equation like F2 = k.F1