- #1

- 29

- 0

## Homework Statement

I have been able to complete section a and b but c doesn't match the answer

a)The acceleration of free fall at the surface of a planet is g and the radius of the planet is R. Deduce that the period of a satellite in a very low orbit is given by T=2pi sqrt(R/g).

b)Given that g=4.5ms^-2 and R=3.4x10^6m, deduce that the orbital period of the low orbit is about 91 minutes.

c)A spacecraft in orbit around this planet has a period of 140 minutes. Deduce the height of the spacecraft from the surface of the planet. (Ans: 3.1x10^6)

## Homework Equations

T=2pi sqrt(R/g)

T=2pi sqrt(R^3/GM)

## The Attempt at a Solution

I can't obtain the answer 3.1x10^6 in c). When doing c), T=2pi sqrt(R/g) can't be used with g=4.5ms^-2 because period of 140 minutes means acceleration of free fall has already changed so I used T=2pi sqrt(R^3/GM):

140x60=2pi sqrt((R+h)^3/GM) {used R+h because the question wants height of spacecraft from SURFACE of planet}. Now I must find M. Using information obtained from b): 91x60=2pi sqrt((3.4x10^6)^3/GM), M=7.81x10^23. Substituting M into 140x60=2pi sqrt((R+h)^3/GM), I get R+h=4.53x10^6, h=1.13x10^6!!??

I've thought this over for a very long time. Have I misunderstood the question or is my method wrong? Someone help me! Thank you!