# Newton's law of universal gravitation

1. Feb 2, 2008

### chocolatelover

Hi everyone,

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

Newton's law of universal gravitation is represented by the following equation where F is the magnitude of the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance.
F = GMm/r2
Force has the SI units kg·m/s2. What are the SI units of the proportionality constant G?

2. Relevant equations

3. The attempt at a solution

Would it be m/kg s^2?

Thank you very much

2. Feb 2, 2008

### Littlepig

Well, use dimensional analysis: in classical mechanics, primary units are Mass, Time, and Space.
You know the dimensions of F, M,m,and r, so, that means you have only 1 var with 1 equation: linear determinate.

Well, in your attempt you say m/kg?!, SI units of m are Kg, so, m/kg is meaningless(and it's not correct), try google something about dimensional analysis, to see how it works. dimensional analysis is SUPER helpful in alot of situations..:P

Littlepig

3. Feb 2, 2008

### mjsd

unit on LHS must equal the unit on the RHS for the equation to make sense. So altogether GMm/r^2 must has unit of Force: Kg. m/s^2... and you can now work out unit G must be

4. Feb 3, 2008

### chocolatelover

Thank you very much

I still don't really understand it. I tried doing cross multiplication and I got G=(kgm)(r^2)/Mms^2 or Kgr^2/Ms^2. Is that anywhere close?

5. Feb 3, 2008

### rock.freak667

$$G=\frac{Fr^2}{Mm}$$

Unit of Force(F) is ___? (and now in SI units)
Unit of distance(r) is ___?
Unit of Mass(M,m) is ___?

When you get those 3 replace them into the equation and simplify.

6. Feb 3, 2008

### chocolatelover

Thank you very much

This is what I did:

I new that the distance had to be in m, so r^2=m^2. I also knew that the mass was kg/m^3 and the force was kgm/s^2. Do I then just need to simplify this?

G=(kgm/s^2)(m^2)/(kg/m^3)(kg/m^3)

Does that look right?

thank you very much

7. Feb 3, 2008

### rock.freak667

Why did you put mass as kg/m^3? Isn't mass just kg?

The force is correct and r is correct. but what you put as mass is off

8. Feb 4, 2008

### chocolatelover

Thank you very much

Since I have M and m, would it be (kg)(kg)?

So, G=(kg/m^3)(m^2)/(kg)(kg)
G=1/mkg

Does that look right?

Thank you very much

9. Feb 4, 2008

### rock.freak667

Yes M and m would be (kg)(kg)

G=(kgms^-2)(m^2)/(kg)(kg)

now simplify that

10. Feb 4, 2008

### chocolatelover

Is it m^3/kgs^2? I don't understand how you got s^-2 or 1/s^2? Could you please explain that to me?

Thank you very much

11. Feb 4, 2008

### TVP45

This is a constant. Always start with the unit you want in your answer. Then make all the other units cancel by either adding factors in the numerator or denominator. Normal practice is to never simplify constants, but you should check with your instructor to see if he actually wants that.

12. Feb 4, 2008

### chocolatelover

Thank you very much

Regards