# How go u find the force of gravity?

1. Feb 26, 2006

### SimpleHarmonicMotion

I can't seem to get a correct # for the force of gravity on earth. I tried this:

6.673e-11(UGC) * 5.9742e24(mass of earth) * 15(mass of object) / 6378.1(radius of earth)^2

For some reason it gives me -.02867... and i no that can't be correct, what did i do wrong><?

2. Feb 26, 2006

### Hootenanny

Staff Emeritus

3. Feb 26, 2006

### Hootenanny

Staff Emeritus
And do you want the force exerted on the object or the gravitational field strength (g)? From your working above your calculating the force exerted on an object of 15kg mass, not acceleration due to gravity ($9.81 ms^{-2}$).

Last edited: Feb 26, 2006
4. Feb 26, 2006

### d_leet

How did you get a negative number? None of the numbers in your expression are negative, so there is no possible way to get a negative answer.. What exactly are you trying to find?

5. Feb 26, 2006

### SimpleHarmonicMotion

huh? I'm confused, how do i find the force of gravity? and what's the radius of the earth?

6. Feb 26, 2006

### Hootenanny

Staff Emeritus
The radius of the earth is wrong you've quoted it in km it should be in metres, the gravitation constant should be negative.

7. Feb 26, 2006

### d_leet

No it shouldn't it's positive 6.67x10-11.

8. Feb 26, 2006

### Hootenanny

Staff Emeritus
Im sorry, I mean there should be a negative sign infront of the equation, the eqaution is usually quoted as:
$$F = - \frac{GMm}{r^2}$$

9. Feb 26, 2006

### d_leet

Hmm... Ok I've never seen it like that, I've never seen it with the negative sign there, but I guess it really doesn't matter as long as you deal with the directions correctly in the problem.

10. Feb 26, 2006

### Hootenanny

Staff Emeritus
That's how I've always used it, but yeah it's irrelevant as long as you define your cordinate system.

11. Feb 26, 2006

### SimpleHarmonicMotion

well, i'm getting 2.86758e-8 but does that make sense? Is this measuring in m/s or N?

12. Feb 26, 2006

### d_leet

Well since you're looking for the force of gravity, what do you think the units should be? And you're number is way off, if you're trying to figure out the force on a 15 kg object on the surface of the earth using the law of gravitation it should be pretty close to 15*9.8

13. Feb 27, 2006

### SimpleHarmonicMotion

well, since i'm way off could u point me in the right direction and tell me what i did wrong? Also, should the units be Newtons since that is the force being applied to the object?

14. Feb 27, 2006

### finchie_88

Looks to me as though you can't use a calculator. the only thing that was wrong with your initial working out is the distance should be in metres not km, other than that, putting numbers into your equation is your only other problem.:rofl:

Its a force, what are the units of force? Do you define a force in terms of $$ms^{-2}$$? What exactly do you think $$ms^{-2}$$ means, what do the letters actually represent?

Last edited by a moderator: Feb 27, 2006
15. Feb 27, 2006

### SimpleHarmonicMotion

6.673e-11 * 2.9742e24 * 15
--------------------------- =
6378000.1^2

6.673e-11 * 2.9742e24 * 15
--------------------------- =
4.0679e13

5.9799e13
--------------------------- =
4.0679e13

1.47 N

Is that correct? And btw, mass of earth is in kg, so does it matter if i put the mass of the object in kg?

16. Feb 27, 2006

### Hootenanny

Staff Emeritus
You can check you answer using $F = ma$. You know that the acceleration due to gravity at the earth's surface is roughly $9.8 m\cdot s^{-2}$, you also know the mass of the object (15kg) and you've calculated a force so;
$$F = ma \Rightarrow a = \frac{F}{m} = \frac{1.47}{15} = 0.098$$
Which means your a factor of $10^2$ out. Check you units, you logic is fine. Yes, mass units must be consistant throughout. If the object is 15g, not 15kg, this would explain yourproblem.

17. Feb 27, 2006

### Eivind

Gravitational acceleration acting on a body at the earth`s surface: (6.67e-11*5.974e24 kg)/(6.4e6)^2 m= 9.73 m/s^2

9.73 m/ss*15 kg=145,95 N

I got an lower answer then 9,81 m/s, but I think it is beacause of some wrong values.

18. Feb 27, 2006

### topsquark

If you are using MKS units, then the mass of both the Earth and object will be in kg.