What is the Period of the Hubble Space Telescope's Orbit?

In summary, the Hubble Space Telescope orbits 614 km above Earth's surface and the period of its orbit can be calculated using the equation T = (2*pi*radius)/velocity. To find the velocity, the centripetal acceleration and the universal gravitational constant can be used, along with the mass of the Earth and the distance between the satellite and Earth. Alternatively, Kepler's third law can also be used to calculate the period.
  • #1
JJones_86
72
0

Homework Statement



The Hubble Space Telescope orbits Earth 614 km above Earth's surface. What is the period of the telescope's orbit?

Homework Equations


T=(2*pi*radius)/v


The Attempt at a Solution



No attempt, Using that equation, I'd have to know the velocity, and that is where I am stuck.. Any help is appreciated!
 
Physics news on Phys.org
  • #2
The centripetal acceleration is a = v^2/r...all you need to know is Earth's mass to solve for the acceleration. Then you could find the velocity.
 
  • #3
Ok, I have the mass of the Earth = 5.9742 × 10^24 kilograms, but where do I plug that in to?
 
  • #4
F = G*Me*Ms/R^2 = Ms*a, Me is mass of Earth and Ms is mass of satellite
a = G*Me/R^2...so for the acceleration you need the mass of the Earth and the radial distance, which is R = Re + h, Re being the radius of the Earth and h being the altitude above Earth's surface (614 km in this case).
Once you have a and R, using a = v^2/R, you could find the velocity
 
  • #5
Ok, but the mass of the satellite is not given...
 
  • #6
No need for the mass of the satellite...it cancels so that a = G*Me/R^2.
 
  • #7
what is the capital G, is it gravity coefficient (9.81 m/s^2)?
 
  • #8
No, its the universal gravitational constant...G = 6.673 x 10^-11 N*m^2/kg^2. If you haven't learned this yet then I guess we'd have to solve this another way.
 
  • #9
No, I haven't learned that yet. You see my professor is a jackass. He's atleast 65 years old, and he is so far behind in lectures as compared to our homework, and we mention this to him and he don't care, all he cares about is his paycheck, so I am stuck trying to learn stuff on my own. None of this stuff is in my book either, which is a E-Book.
 
  • #10
Using the information you gave me I am still getting the wrong answer. I need the final answer in hours, I am so confused.
 
  • #11
oh...well...that makes things a bit complicated. I'll go ahead and scratch a bit of the surface of what we know about gravity for you. Gravity is a field force that occurs between masses. To calculate the force of gravity between masses, the general classical expression for is F = G*m1*m2/r^2, m1 and m2 being the 2 masses and r being the distance between their centers. The value of this force is both the force on m1 and m2 and it is an attractive force. In this case, the planet Earth is exerting a force on the satellite. At the same time, the satellite has a tangential velocity, which is what you're looking for, that is keeping it from heading straight towards the Earth (instead, it orbits the planet). So in this case, the force is point toward the center of the satellite's orbit, making the force a centripetal force...but I'm not sure if you're supposed to know this or not.

what answer are you getting (so that I can check it with what I have)?
 
  • #12
Thanks for that information. I found something related to a similar problem using Kepler's third law, are you aware of what this i and how I could apply it to this problem?
 
  • #13
Kepler came up with his laws empirically before Newton came up with his work. Decades after Kepler's work, Newton did something pretty amazing, which was to prove Kepler's laws with his own work...this further proved the accuracy of his findings and Kepler's. Altogether, using Kepler's laws would get you the same answer as with Newton's. If you were to use the equation for Kepler's third law, all you'd need would be Earth's mass and the distance between the satellite and earth, which is Re + h. With that you could find the period.
 
Last edited:
  • #14
Ok, I am getting nowhere, I know the answer is supposed to be around 1.6 hrs, can you give me a few extra tips? Thanks!
 
  • #15
Alright...that seems to be the right answer.
The force between the satellite and Earth is F = G*Me*Ms/(R^2)...it is the force on the satellite and on the earth. Me is the mass of the earth, Ms is the mass of the satellite, and R is the distance between the two (which is the addition of the radius of the Earth and the altitude of the satellite above the Earth's surface...so you'll need to know the radius of the earth...the problem gives you the altitude, just be sure to convert it to meters). Force is also equal to M*a...so in the case of the satellite, F = G*Me*Ms/(R^2) = Ms*a. The Ms cancels so that a = G*Me/(R^2), so you do not need the mass of the satellite. Once you have a (and R)... you could solve for v in a = (v^2)/R. The period is T = 2*pi*R/v...the answer you'll get will be in seconds...you'll need to convert to hours.
 
  • #16
OOOk, I've finally figured it out using Kepler's third law, Thank you for your help Gear300, I really do appreciate it.
 
  • #17
alright then...looks like you did it differently...as long as you understand the concept of things, you'll be alright.
 

FAQ: What is the Period of the Hubble Space Telescope's Orbit?

What is a circular orbit?

A circular orbit is the path an object takes around another object in space, where the distance between the two objects remains constant.

How is the period of a circular orbit calculated?

The period of a circular orbit is calculated by using the equation T = 2π√(r³/GM), where T is the period, r is the distance between the two objects, G is the gravitational constant, and M is the mass of the larger object.

What factors affect the period of a circular orbit?

The period of a circular orbit is affected by the distance between the two objects, the mass of the larger object, and the gravitational constant.

Can objects have different periods in the same circular orbit?

No, all objects in a circular orbit around the same object will have the same period, regardless of their mass or distance from the larger object.

How does the period of a circular orbit change as the distance between the objects increases?

The period of a circular orbit increases as the distance between the two objects increases. This is because the gravitational force between the objects decreases as the distance increases, leading to a longer period.

Similar threads

Back
Top