# Ratio of the Magnitude of gravitational force?

1. Sep 29, 2015

### Albert24

1. The problem statement, all variables and given/known data
Calculate the magnitude of the gravitational force between the earth and a 3.7 kg mass on the surface of the earth. The distance to the center of the earth from the surface is 6370 km and the mass of the earth is 5.98·1024.
That gave me 36.37 N

Calculate the magnitude of the gravitational force between the moon and a 3.7 kg mass on the surface of the earth nearest to the moon. The distance to the center of the moon from the surface of the earth is 376,000 km and the mass of the moon is 7.36·1022 kg.
That gave me 1.285×10-4 N

Calculate the ratio of the magnitude of the gravitational force between a 3.7 kg mass on the surface of the earth due to the sun to that due to the moon. The mass of the sun is 1.99·1030 kg and the distance from the center of the sun to the surface of the earth is 1.50·108 km. ?

2. Relevant equations

F=GMm/r^2

3. The attempt at a solution
I dont quite understand the problem, is it asking what is the magnitude of the gravitonial force between the mass and the sun taking in count the gravitational force between sun anf the moon? In that case the ratio=
GM1m2/r^2/GM1m2/r^2 ??

2. Sep 29, 2015

### SteamKing

Staff Emeritus
No, the problem is asking what is the gravitational force between the sun and a 3.7 kg mass on the surface of the earth. Call this force Fsun.
You calculated the gravitational force between this same object on the earth's surface due to the moon, Fmoon, to be 1.285×10-4 N.

What is the ratio Fsun / Fmoon ?

3. Sep 29, 2015

### Albert24

Thanks it was 171 N

4. Sep 29, 2015

### SteamKing

Staff Emeritus
Can't be.

1. A ratio of two forces has no units.
2. You're saying that the force exerted by the sun on an object sitting on the earth is orders of magnitude greater than what earth's gravity exerts on that same object.
Do you think that's reasonable?

5. Sep 29, 2015

### Staff: Mentor

It's the ratio of the Sun's force to the Moon's force on an object located near the surface of the Earth.

6. Sep 29, 2015

### Albert24

I am sorry you are right there are no units they cancel out, since I was working with forces I had Newton in mind