Calculating Period of Hubble Telescope Orbit

• cstout
In summary, the Hubble Space Telescope orbits Earth at an altitude of 615 km and the period of its orbit can be calculated using the equation T = 2(pi)r/v, where r is the radius of the Earth plus the altitude. The approximate period of the telescope's orbit is 1.61 hours or 96.85 minutes. Alternatively, the period can also be calculated using Kepler's 3rd law, which states that P^2 = (4(pi)^2)r^3/(M+m)G, where r is the distance between the centers of mass of the Earth and the telescope, M is the mass of the Earth, m is the mass of the telescope, and G is the gravitational constant. Using this
cstout

Homework Statement

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

The Attempt at a Solution

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

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

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

cstout said:
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?

it orbit 615km above the Earths surface

The orbital radius is the distance from the centre of the Earth to the satelite.
Hence the radius of the Earth PLUS the altitude = approx 7000km.

Remember to express total distance (Earth's radium + orbital distance) in meters to mantain a coherence among unities.

And since you aren't given the speed you also need Kepler's 3rd law to work out the period. Not sure where you got the 9.8 from in your original equation - but it's wrong!

how would you use Kepler's 3rd law in relation to this question though, you aren't given some of the information needed.

Nope - you have everything you need. What information do you think you need for Kepler's 3rd Law that you aren't given?

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.

Remember the simplified form of the law doesn't need the mass of the orbital body
, you can assume that G(M+m) is pretty much equal to GM. Hubble only weighs about 10tons!

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

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

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.

Last edited:
I don't think so..

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

What is a circular orbit?

A circular orbit is the path that an object follows around another object, where the distance between the two remains constant and the object travels at a constant speed.

What is the period of a circular orbit?

The period of a circular orbit is the time it takes for an object to complete one full revolution around another object.

What factors affect the period of a circular orbit?

The period of a circular orbit is affected by the mass of the objects, the distance between them, and the gravitational force between them.

Can the period of a circular orbit be changed?

Yes, the period of a circular orbit can be changed by altering the speed or distance of the orbiting object, or by changing the mass of the objects.

How is the period of a circular orbit calculated?

The period of a circular orbit can be calculated using the formula T = 2π√(r³/GM), where T is the period, r is the distance between the objects, G is the gravitational constant, and M is the combined mass of the objects.

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