Mad scientist istalled a cannon on the North pole, and tried to shoot down all geosyncronous satellites?

Question

With threat effort the scientist finally managed to achieve minimum muzzle velocity necessary to success of his mission.
After nearly missing one satellite the cannon ball landed on the South pole.

How much time did the cannon ball spend in space?
(Assume that radius of the GSS orbit is large compared to radius of Earth)

Oh, I forgot.
There are 24 hours in one day

Answer 1

this question doesn't have enough clues. like how fast the cannon ball was going at. and what time did the scientist shoot the cannon ball at and what time it landed at the south pole.

Answer 2

12 hours!?

The period of an orbit is a function of G*M (gravitational constant x mass of earth) and the major axis length.

Assuming the cannonball just missed a satellite at its apogee, then the major axis lengths are the same. And since GM is the same for both the cannonball and the satellite, then the periods must be the same. In this case, the time the cannonball spent in orbit is 1/2 the period = 24hr/2 = 12 hours!

Note I basically ignore the time spent between the ground and "space" since this is much much smaller than the distance it travels in space, so it will be like a few minutes out of a few hours, so its negligible.Just treat it as an orbital problem and ignore the atmosphere.

Answer 3

He tells you how high it goes by saying it nearly missed a geosynchronous satellite at about 22,000 miles high. The fact that it lands on the south pole means it is traveling an eliptical path. A minimum eliptical escape velocity is 22,000 MPH. You know the distance from the north to south pole.

Answer 4

12 hours, assuming the satellite was in equitorial geosynchronicity. (i.e., geostationary) and the equitorial circumference of the earth is the same as the polar circumference.
Falling objects fall to the earth at the same rate regardless of tangental velocity.
Orbiting objects are falling around the earth, and geosynchronous objects fall completely around the earth in one day.
When it barely missed the satellite, the cannonball was at the same height. Since it takes 6 hours for the satellite to go 1/4 of the way around the earth, it took the cannonball the same to land. (Plus 6 hours to go up, since the parabolic trajectory of the cannonball is symmetric on both sides of the maximum height.)

Answer 5

One would have to know the weight of the cannon ball, the time of day, how "high", the initial velocity, correolis effect, the tilt of the earth, the angle it was shot at, etc.
By the way, what angle was it shot at?