Question

Scientists want to place a 4100 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.4 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:
mmars = 6.4191 x 1023 kg
rmars = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2

What should the radius of the orbit be (measured from the center of Mars), if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

Answer 1

Any takers on helping me with this problem?

Answer 2

I know that there are two parts to this question...one of them would be finding the gravitational attraction between the two masses, which would use the equation: \[F_{g}=G \frac{ m_{1}m_{2} }{ r^{2} }\] The second part would be to calculate your orbital velocity; sadly, I do not know how to do that.

Answer 3

Ok, from the problem situation, it tells you that the minimum radius for the satellite to orbit mars is 2.4x the radius of Mars above its surface. When an object achieves orbital velocity, its revolution takes just as much time as the planet itself. If you want the revolution to take 8x as long as the planet's revolution, you multiply that 2.4x radius of mars by 8, which means that the satellite would need to be 19.2x the radius of mars above mars' surface to take 8x longer to go around the planet...
I do not think it would be that simple, but it's a decent guess.

Answer 4

I found the attraction to be 1316.8N in a previous problem.

Answer 5

What speed should the satellite have to be in a perfectly circular orbit?
1926m/s
How much time does it take the satellite to complete one revolution?
10.5hrs

Answer 6

Just by multiplying the r by 19.2 it gave me 65,222,400m and it was wrong...

Answer 7

Does anyone else have any ideas on this one? I haven't gotten anything to work yet.

Answer 8

You could try to use trigonometry on this equation. You have 1,926m/s and 10.5hrs. After 10.5hrs at the previous velocity, you would have completed a revolution on a fixed orbit. 72,802,800m would be the distance traveled around the planet, or the circumference. Divide this by 2pi, 11586,,925.49m would be the radius of the circumference. Subtract the 3,397,000m, the radius of mars, from the radius of the entire orbit. 8,189,925.49m is the distance between the planet surface and the orbit. If you want the satellite to orbit 8x slower, you could multiply the circumference of the orbit times 8. 92,695,403.93m is the radius of the new orbit, then you subtract the radius of mars from that orbit. The distance between the surface of mars and this new orbit would be 89,298,403.93m, or 89,298.40393km. This may not be exact because I did not include all the factors such as gravitational attraction with an increased radius, but it is a much closer approximation.

Answer 9

geometry, not trigonometry...was thinking about radians and arcs when i started typing

Answer 10

Hmmm....still not working. My tutor is even struggling to figure this problem out....