# The Hubble Space Telescope orbits Earth 615 km above Earth's surface.?

The Hubble Space Telescope orbits Earth 615 km above Earth's surface. What is the period of the telescope's orbit?

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- DHLv 710 years agoBest Answer
From Newton's Laws we have

GMm/r^2 = m*v^2/r but v = 2πr/T

So GM/r^2 = 4π^2r/T^2

So T = sqrt(4π^2*r^3/GM) = sqrt(4π^2*(6.38x10^6 + 615x10^3)^3/(6.67x10^-11*5.98x10^24)

= 5820s = 1.62 hr

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