Exam review question about space shuttle orbits at differing radii

[npersons274]

Homework Statement

The space shuttle makes 1 revolution around the Earth in 1.5 hours when it is in an orbit 200 km above the Earth’s surface. The radius of the Earth Re is 6.5 × 10^6m. If the shuttle moves to a new orbit such that it makes 1 revolution per day (24 hours), what is the radius of the new orbit?

(1) 6.2Re (2) 12Re (3) 24.8Re (4) 16Re (5) 0.38Re

Homework Equations

Force_centripetal=Force_gravity

v=(2piR)/T

T=period

The Attempt at a Solution

My professor said:

"Yes, the centripetal force is provided by gravity. The tangential velocity v=2piR/T where T is the period. Thus show that T is proportional to R^2 and solve. For orbit 2, R_2=R_1sqrt(T_2/T_1)"

...but I still can't understand what he's talking about. What does he mean that period T is proprtional to r^2? And what does any of this have to do with balancing the centripetal force with the force of gravity?

Thank you. This forum is the best.

 

[tiny-tim]

welcome to pf!

hi npersons274! welcome to pf!

(try using the X² and X₂ buttons just above the Reply box )

"Yes, the centripetal force is provided by gravity. The tangential velocity v=2piR/T where T is the period. Thus show that T is proportional to R^2 and solve. For orbit 2, R_2=R_1sqrt(T_2/T_1)"

...but I still can't understand what he's talking about. What does he mean that period T is proprtional to r^2? And what does any of this have to do with balancing the centripetal force with the force of gravity?

hint: what is the equation balancing the centripetal acceleration with the gravitational acceleration?