# Derivation of mgh

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1. Dec 14, 2016

### Faiq

1. The problem statement, all variables and given/known data
Derive E=mgh from Newton's law of Gravitation where h is very small. (Use binomial expansion)

2. The attempt at a solution
$$E = \frac{GMm}{(r+h)^2}-\frac{GMm}{r^2}$$
$$E = \frac{GMm}{r^2}(\frac{1}{(1+\frac{h}{r})^2}-1)$$
$$E = \frac{GMm}{r^2}((1+\frac{h}{r})^{-2}-1)$$
$$E = \frac{GMm}{r^2}(1+\frac{-2h}{r}-1)$$ Other powers of h/r becomes negligible for h<<r
$$E = \frac{GMm}{r^2}(\frac{-2h}{r})$$

Not sure where I went wrong or what to do next
Note:- I know this can be solved in a million other methods but I want the answer specifically from this method. I had a book which used this method and I cant remember how this method works out.

2. Dec 14, 2016

### Orodruin

Staff Emeritus
The potential in Newtonian gravity is negative. You are taking potential(reference level) - potential(height h) instead of the other way around.

3. Dec 14, 2016

### Faiq

Okay so I will have 2h/r instead of -2h/r. Doesn't solve the problem though

4. Dec 14, 2016

### Orodruin

Staff Emeritus
Then what is your problem? It looks fine to me.

5. Dec 14, 2016

### Faiq

Instead of "2h/r", I should get "h" so I can rewrite the answer as mgh

6. Dec 14, 2016

### Orodruin

Staff Emeritus
Move 2/r from the h/r term to the term with G and M ...

7. Dec 14, 2016

### Faiq

so 1/r^2 will become 1/r^3.

8. Dec 14, 2016

### Staff: Mentor

Start with gravitational PE, not force. (What you are calling "E" is force not energy.)

9. Dec 14, 2016

### Faiq

ohhhhhhhhhh

10. Dec 14, 2016

### Faiq

Okay now I am getting GMm/r *h/r

11. Dec 14, 2016

### Faiq

Oh okay done thanks

12. Dec 14, 2016

### Orodruin

Staff Emeritus
True that. The same basic principle applies though. The entire point is expressing g in terms of G, M and r.

13. Dec 14, 2016

Yep.