Question

A spy satellite completes six revolutions every day in a circular orbit. Determine the altitude of the satellite above Earth's surface. the speed of the satellite.

Answer 1

time period of the satellite=24/6=4 hours

hence T=4 hours=4*3600 seconds=14400 seconds

let speed of the satellite be v and radius of orbit be R.

then time period=T=2*pi*R/v...(1)

balancing centripetal force with gravitational force:

m*v^2/R=G*M*m/R^2

==>v^2=G*M/R...(2)

using v=2*pi*R/T

then 4*pi^2*R^2/T^2=G*M/R

==>R=(G*M*T^2/(4*pi^2))^(1/3)

using the values,

R=(6.674*10^(-11)*5.972*10^24*14400^2/(4*pi^2))^(1/3)

=1.2793*10^7 m

so height above earth's surface=R-radius of earth

=6.4215*10^6 m

=6421.5 km

part b:

v=2*pi*R/T

=2801.9 m/s