Q: In Kepler's Laws "radius" means the angular-averaged radius, but what's the time-averaged radius?

A: (1 + 0.5 `e`^{2}) `a` where
`e` = eccentricity, `a` = semi-major axis.

Q: What are the speed and period of a satellite orbiting at the earth's surface?

A: Assume the earth has mass 6e24 kg and radius 6.4e6 m. Then the
speed of a satellite orbiting the earth at its surface is

`V` = 7940 m/s = 17800 mph

and its orbital period is

`P` = 5070 s = 84.4 min.

Escape velocity at the earth's surface is

`E` = sqrt(2) `V` = 11200 m/s = 25200 mph.

Some more facts about orbits:

If you dig a straight tunnel between *any* two points A and B
on the earth's surface and drop a vehicle down it, it will oscillate
between the ends of the tunnel with period `P`. If the tunnel
passes through the earth's centre O then its velocity there will be
`V`. If the acceleration of the vehicle is magically reversed
between O and B then it will have speed `E` at B.

The time taken to fall through a tunnel in a planet of given density is independent of the radius, by dimensional analysis alone [M. Pilloff, rec.puzzles].

`P` is also the period of a Shuler pendulum, ie a pendulum
of length the radius of the earth whose bob is at the earth's
surface. If the bob is released from the same height as the pivot it
will pass through its lowest point with speed `V`.

Q: If a (light) satellite in a circular orbit is suddenly stopped, how long does it take to hit the (point) mass it's orbiting?

A: 1 / (4 sqrt(2)) = 17.7% of its orbital period, so for example 4h 15m for a geostationary satellite, 4d 19h for the moon. The time doesn't depend on anything else.

This page is maintained by Thomas Bending,
and was last modified on 7 March 2017.

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Copyright © Thomas Bending 2017.