Determine the distance of this point from the center of the Earth.

[Sheneron]

Homework Statement

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?

Homework Equations

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

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 each other and solve for h

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

is this right so far?

 

[Hootenanny]

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.

 

[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?

 

[Hootenanny]

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

Are both solutions physically possible?

 

[Sheneron]

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

 

[Hootenanny]

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

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

 

[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)

 

[Sheneron]

hootenanny? anybody?

 

[dwahler]

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

 

[Sheneron]

They are both between the Earth and the moon.

 

[dwahler]

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


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

 

[Sheneron]

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