# Homework Help: Determine the distance of this point from the center of the Earth.

1. Apr 15, 2008

### Sheneron

1. The problem statement, all variables and given/known data
On the way to the moon the apollo astronauts reached a point where the Moon's gravitational pull became stronger than the Earth's.
A) Determine the distance of this point from the center of the Earth.
B)What is the acceleration due to the Earth's gravitation at this point?

2. Relevant equations
$$F_G = \frac{Gm_1m_2}{r^2}$$

3. The attempt at a solution
I tried to solve this but am not sure if I am doing it right:
d= distance from earth to moon
h= height above the earth
$$g_{earth} = \frac{GM_E}{h^2}$$
$$g_{moon} = \frac{GM_M}{(d-h)^2}$$

set the equations equal to eachother and solve for h

$$d^2 - 2dh + h^2 = \frac{M_m}{M_E}h^2$$

is this right so far?

2. Apr 15, 2008

### Hootenanny

Staff Emeritus
Your good so far with the exception that h is the distance from the centre of the earth, not the height above the earth's surface.

3. Apr 15, 2008

### Sheneron

I think i have it. I am solving it as a quadratic, so how do I know which of the two answers it is?

4. Apr 15, 2008

### Hootenanny

Staff Emeritus
Are both solutions physically possible?

5. Apr 15, 2008

### Sheneron

indeed, very. Would you like me to post what I got?

6. Apr 15, 2008

### Hootenanny

Staff Emeritus
If you post what you've got I'll have a good look at it.

7. Apr 15, 2008

### Sheneron

M_e = 5.98 x 10^24
M_m = 7.36 x 10^22
d= 3.84 x 10^8

$$0.98769h^2 - 7.68 x 10^8h - 1.475 x 10^{17} = 0$$

I put that all in the quadratic formula.. I can't seem to get it to work on here but this is what I got:

7.68 x 10^8 +/- 8.42 x 10 ^7 all over 2(0.98769)

Last edited: Apr 15, 2008
8. Apr 15, 2008

### Sheneron

hootenanny? anybody?

9. Apr 15, 2008

### dwahler

Try calculating the actual distances for both roots. Would you encounter both of them on a trip between the earth and the moon?

10. Apr 15, 2008

### Sheneron

They are both between the earth and the moon.

11. Apr 15, 2008

### dwahler

Check again. Based on your equations so far, I get:

$$h_1 = 3.46 \cdot 10^8 \ \mathrm{m}$$
$$d = 3.84 \cdot 10^8 \ \mathrm{m}$$
$$h_2 = 4.31 \cdot 10^8 \ \mathrm{m}$$

12. Apr 15, 2008

### Sheneron

Oooo. Thanks for the help. I got it now.