You have been sent in a small spacecraft to rendezvous with a space station that

Question

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius 2.5500·104 km from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position, when the space station has come halfway around the circle to meet you. Calculate the minimum radius from the Earth's center - the low point - of your new elliptical orbit. Is it greater than the radius of the Earth (6370 km), or have you botched your last physics problem?

Explanation / Answer

Kepler's 3rd law says that T^2 is prop to a^3, where a is the semi-major axis of an elliptical orbit. For a circular orbit, this is just the radius.

In your case then we are given that Te = (1/2) Tc (Te is for elliptical orbit, Tc is for circular) and

Te^2 = (1/4) Tc^2

So...

a^3 = (1/4) r^3 where a is for the elliptical orbit and r is the circular

take cube root

a = 0.62996 * r = 0.62996 * 25500 km =

= 16064 km

This means the major axis (from high point to low point in orbit) is twice this value, or 32128 km.

Which means that since the high point is 25500 km from the center of Earth, the low point is

32128 - 25500 = 6628 km from the center of Earth.

Since the radius of Earth is about 6380 km, you'll just barely make it... the low point is about 248 km above the surface!

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