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## Homework Statement

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

## Homework Equations

T = 2(pi)r/v

## The Attempt at a Solution

T = 2(pi)(6.37x10^6)/9.8

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- Thread starter cstout
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- #1

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

T = 2(pi)r/v

T = 2(pi)(6.37x10^6)/9.8

- #2

mgb_phys

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Remember the radius of the orbit is measured from the centre of the earth.

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Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.

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Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.

So, the radius of the orbit is radius of earth? That means the satellite is orbiting in surface, right?

- #5

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it orbit 615km above the Earths surface

- #6

mgb_phys

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Hence the radius of the Earth PLUS the altitude = approx 7000km.

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- #8

mgb_phys

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And by the way, mgb_phys - I'll be cstout was using the value for the approximate acceleration due to gravity at the Earth's surface, i.e. 9.8 m/s^2, which you'd just have to call a pretty bad guess.

- #11

mgb_phys

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, you can assume that G(M+m) is pretty much equal to GM. Hubble only weighs about 10tons!

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Ok, I finally got it, the answer is 1.61 hrs. Thanks for the help

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mgb_phys

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That sounds about right, 90 minutes is pretty typical for LEO

- #14

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Try using this formula: 4(pi)² r³/(M+m)G = P²

Where:

r = radius between the centers of mass of both bodies measured in meters

M = the larger mass in kilograms (5.9742 X 10^24 for the earth)

m = the smaller mass in kilograms (this mass is usually so much smaller than the

larger mass that it can almost always be disregarded)

G = 6.67259 X 10-¹¹

P = the period of orbit in seconds

This formula yields an answer of approx. 5,811 seconds, which is about 96.85 minutes.

Where:

r = radius between the centers of mass of both bodies measured in meters

M = the larger mass in kilograms (5.9742 X 10^24 for the earth)

m = the smaller mass in kilograms (this mass is usually so much smaller than the

larger mass that it can almost always be disregarded)

G = 6.67259 X 10-¹¹

P = the period of orbit in seconds

This formula yields an answer of approx. 5,811 seconds, which is about 96.85 minutes.

Last edited:

- #15

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Since the radius,r of the telescope is just about 10% of the radius of the Earth.

Then it can be ignored..

Just use R for the radius of Earth.

Hence, Period of revolution of telescope

= 2Pi Square root R/g

=84.43min

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