### Re: What is the speed of most asteroids in mph?

Date: Wed Mar 21 04:22:48 2001
Posted By: Chris Lintott, Undergraduate, Astrophysics, Magdalene College
Area of science: Astronomy
ID: 985108871.As
Message:
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I don't know the answer to your question - but if we make some simple
assumptions then it's easy to work out. If we assume that the asteroids
are travelling in circular orbits around the Sun (this is pretty good for
those in the main asteroid belt) then we can say the gravitational force
is given by the formula

F = m x v squared divided by the radius. That's true for anything
travelling in a circle; m is the mass of the asteroid and v is its speed.

Newton said that the force was also equal to

F= G M m divided by radius squared. M is the mass of the Sun, m is the
mass of the asteroid and G is a constant, equal to about 0.00000000007
(Discovering what G is has taken many years...but that's another story).

Rearranging these gives a formula for the speed; the Sun weighs 2 x 10^30
kg - that's 2 followed by 30 zeroes! A typical radius for an asteroid in
the main belt is 300 000 000 km (about twice the Earth's distance from the
Sun)  Plugging all these numbers in gives me a value of v = 21 km per
second. Converting to miles per hour gives about 47000 mph.

These values do pretty well for the asteroids in the main belt. What about
those whose orbits swing them close to the Sun and then send them far out
again? Well, this value is a pretty good average; but you have to remember
that the closer they are to the Sun, the faster they move - its force on
them is greater and so they will speed up. This law was discovered by
Kepler. In the same way, planets closer to the Sun orbit faster - so Earth
is travelling faster than Mars

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