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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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