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

AI Thread Summary
The discussion revolves around calculating the distance from the center of the Earth to a point where the Moon's gravitational pull surpasses that of the Earth. The user initially sets up equations for gravitational forces from both the Earth and the Moon, aiming to find the height above the Earth's surface where this occurs. Clarifications are made regarding the definition of variables, particularly that 'h' represents the distance from the Earth's center. The user successfully derives a quadratic equation and seeks assistance in solving it, ultimately finding two possible solutions for 'h'. The final distances indicate that both solutions are feasible within the context of a trip between the Earth and the Moon.
Sheneron
Messages
360
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I think i have it. I am solving it as a quadratic, so how do I know which of the two answers it is?
 
Sheneron said:
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?
 
indeed, very. Would you like me to post what I got?
 
Sheneron said:
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.
 
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:
hootenanny? anybody?
 
Try calculating the actual distances for both roots. Would you encounter both of them on a trip between the Earth and the moon?
 
  • #10
They are both between the Earth and the moon.
 
  • #11
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
Oooo. Thanks for the help. I got it now.
 
Back
Top