Anonymous

# The Hubble Space Telescope orbits Earth 613 km above Earth's surface.?

What is the Period of the telescopes orbit?

Relevance

Let telescope hav e a mass m and velocity v.

Then "centrifugal force" (or whatever you call it) = m v^2/R

where R = r + 613,000 (r is earth's radius, meters)

This must equal gravitational attraction. We know that at the Earth's surface this would be equal to mg. We also know that it obeys an inverse square law. So at 613 km, it will be equal to mg x r^2/R^2

So mv^2/R = mg x r^2/R^2

m of course cancels out. You need to substitute values of r and g, and do the arithmetic.

Knowing v, period = 2 pi R/v (length of orbit divided by speed)

Make sure you understand each point in the reasoning. the secret is that gravitational force has to be equal to centrifugal "force". the rest is just thinking about the geometry, and tedious but straightforward arithmetic.

Good luck

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