# Space station orbit gravity help

1. Oct 8, 2007

### pinkyjoshi65

A space station is in orbit between the earth and the moon. The force due to gravity on the space station from the moon is the same as the force due to gravity from the earth. How far away from the earth is the speace station? How far from the moon is the space station?

2. Oct 9, 2007

### cristo

Staff Emeritus
What are your thoughts on the question?

3. Oct 9, 2007

### pinkyjoshi65

i tried doing the question, but got stuck in it. I found F_gm and F_ge. Then i equated them. i ended up with r/R=0.11045, where r=dist. between the satelite and the moon, and R=dist. between the satellite and earth.

4. Oct 9, 2007

### learningphysics

Good so far. Now use the distance between the earth and the moon and set this equal to r + R... look up the distance between the earth and the moon.

5. Oct 9, 2007

### pinkyjoshi65

so by doing that, i got the R=346168670.4m and r=38234329.6m

6. Oct 9, 2007

### learningphysics

Looks good to me.

7. Oct 9, 2007

### pinkyjoshi65

i had another question: Mars has 54% the mass of the Earth and a radius 11% of the earth. what is g on mars? what i did for this was
M_m=0.54M_E, R_m=0.11M_E
g=GM_m/R_m^2
g=G*0.54M_E/0.11R_E
g=(6.67*〖10〗^(-11)*0.54*6* 〖10〗^24)/(〖0.11*6.37*〖10〗^6)〗^2
g=21.61*〖10〗^13/0.49*〖10〗^12
g=44.102*10= 441.02m/sq sec
But when i searched the net for the g on mars..i found out tht the g is 3.77m/sq. sec

8. Oct 9, 2007

### malawi_glenn

You must have mixed the radius and mass of mars. Mars has 11% of earths mass, and 54% of earth radius.

You should have looked up that too =)

9. Oct 9, 2007

### Hootenanny

Staff Emeritus
Incedently, you can work this one out without looking up any values. You know that g is proportional to m and inversely proportional to the square of r, hence you can write;

$$g_m = \frac{0.11}{0.54^2}\cdot g_e$$

Last edited: Oct 9, 2007
10. Oct 9, 2007

### pinkyjoshi65

so the question is wrong?

11. Oct 9, 2007

### Hootenanny

Staff Emeritus
The question as written above in post #7 is incorrect. As malawi_glenn correctly states, the mass of mars is approximately 11% of the earth's and mars' radius is approximately 54% of the earth's.

12. Oct 9, 2007

### thiotimoline

Data I used:
Mass of Earth = 6.02 x 10^24 kg = Me
Mass of Moon = 7.34 x 10^22 kg = Mm
Let R = distance between Earth and station, r =distance between Moon and station

At equilibrium,

Force acting on station by Earth = Force acting on station by Moon
GMeM / R^2 = GMmM / r^2
r / R = sqrt (Mm / Me )
= 0.1104

And knowing R + r = distance between Earth and Moon = 3.844 x 10^9 m,
R = 3.367 x 10^9 m
r = 4.77 x 10^8 m

Station is closer to Moon than to Earth.

13. Oct 10, 2007

### pinkyjoshi65

r+R is supposed to be equal to 3.84*10^8m. How did u get r+R =3.84*10^9m??

14. Oct 10, 2007

### Hootenanny

Staff Emeritus
The sources I have agree with 3.84*10^8m.

15. Oct 10, 2007

### pinkyjoshi65

also i got r= 3.82*10^7m not 4.77*10^8m. Which one is right?

16. Oct 10, 2007

### Hootenanny

Staff Emeritus
Forgive me, but I'm not checking through the arithmetic; I'm assuming that if thiotimoline used an incorrect value, his/her final result will be incorrect.

Last edited: Oct 10, 2007
17. Oct 10, 2007

### pinkyjoshi65

Thanks..:)

18. Oct 10, 2007

### Hootenanny

Staff Emeritus
Pleasure

19. Oct 14, 2007

### thiotimoline

This is the correct ans, kindly ignore my previous post. I apologise for the mistake. :)