# Newton's second law and how to calculate the force on the mo

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1. Jan 10, 2016

I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?

2. Jan 10, 2016

### Student100

What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?

3. Jan 10, 2016

Objects falling towards the moon*

4. Jan 10, 2016

### Student100

Simply redefine g. That's all that needs to be done.

5. Jan 10, 2016

Yeah, but what I was asking is there any way to use earths gravity as a template without actually redefining g, which now I think about, I did not specify, I got the formula f=m(g/6). But is there a better way of doing it? My formula skills aren't that great :/

6. Jan 10, 2016

### Student100

Why not use Newton's law of gravitation in that case?

$$F=\frac{Gm_1m_2}{r^2}$$

Although, g in F=mg is just a dummy variable, it doesn't have an intrinsic meaning until we give it one, so I'm not really understanding where you're going.

7. Jan 10, 2016

We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.

8. Jan 10, 2016

### Student100

It doesn't use Earth as a template either, (not sure why you'd want to), it uses the nature of the system and the gravitational constant.

9. Jan 12, 2016

### Domullus

Free fall acceleration on Earth is just $g=G\frac{M_{Earth}}{R^2_{Earth}}$, here $M_{Earth}$ is Earth mass, and $R_{Earth}$ - its radius. For Moon you get $g=G\frac{M_{Moon}}{R^2_{Moon}}$. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation $g_{Moon}/g_{Earth}$. So you will get $g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}$

Last edited: Jan 12, 2016
10. Jan 12, 2016

### sophiecentaur

It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler .

11. Jan 13, 2016