# Gravitational Acceleration Help

1. May 17, 2004

### wikidrox

Show that GMm (1/r1 - 1/r2) and mgh are, for practical purposes, equal, when r2 is almost equal to r1 near the earths surface.

I just don't know how to do it.

2. May 17, 2004

### arildno

1.
$$g=\frac{GM}{r_{1}^{2}}$$, where $$r_{1}$$ is the radius of the earth.

2. Set $$r_{2}=r_{1}+h$$, and assume $$h<<r_{1}$$

3. May 17, 2004

### AKG

Okay, if you're trying to prove that:

$$GMm(\frac{1}{r_{1}}\ -\ \frac{1}{r_{2}})\ =\ mgh$$

Notice that you can start by cancelling out "m" from both sides. Also, notice that you can rewrite the stuff in brackets like so:

$$GM\frac{r_{2}\ -\ r_{1}}{r_{2}r_{1}}\ =\ gh$$

Now, can you think of a relationship between h, r1, and r2? Also, recognize that r1 is the radius of the Earth. Now, can you use the information from the previous sentence to make an assumption so that you can do something with the r1r2 part?

4. May 27, 2004

### wikidrox

hgfdkl;

I still can get it totally.

here is where I get to:

GM/r1^2 + r1h = g

I can't get rid of the r1h.

5. May 27, 2004

### arildno

Let's do this as follows:

1. You are to approximate:
$$GMm(\frac{1}{r_{1}}-\frac{1}{r_{1}+h})$$

2. You know the following:
$$\frac{GM}{r_{1}^{2}}=g,h<<r_{1}$$

3. Bring the difference in 1. together like this:
$$GMm(\frac{1}{r_{1}}-\frac{1}{r_{1}+h})=GMm(\frac{r_{1}+h-r_{1}}{r_{1}(r_{1}+h)})=$$
$$m\frac{GM}{r_{1}^{2}}\frac{h}{1+\frac{h}{r_{1}}}=\frac{mgh}{1+\frac{h}{r_{1}}}$$

4.
Now, by 2:
$$h<<r_{1}\to\frac{h}{r_{1}}<<1$$

The fraction in the denominator is seen to be much less than 1, and therefore we have the approximate equality:
$$\frac{mgh}{1+\frac{h}{r_{1}}}\approx{mgh}$$

6. May 29, 2004

### Gza

Then there's always a series expansion