Time of satellite orbit around moon?

In summary: And that would let you solve it.This is how physics works. You know some stuff, you make some guesses, you see what consequences they have, and you check if they match what you've observed. If they don't match, you go back to making guesses. If they do match, you move on to the next problem.
  • #1
Dillion
39
0

Homework Statement


What is the time of an orbit of a satellite around the moon close to the moon's surface? What quantities do you have to look up to be able to answer this question?

Radius of moon is 1736482 meters

Homework Equations



F = Gm1m2/r^2
and maybe
F=mv^2/r

2pir

The Attempt at a Solution



  1. 2 pi (1736482 meters) = 10910638.19 m = distance around moon
...and that's all I have. I don't know how to achieve time just given a radius. I do know that centripetal force has to come into play somewhere.

Would the answer for the second question just be the radius?

I'm so lost!
 
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  • #2
Hi Dillion.

Let's take it step by step.

For an object to move in circles (e.g. an orbit), what magnitude must the net force acting on it be?
 
  • #3
You have the two key equations. Do you understand them? What does ##F=mv^2/r## mean?
 
  • #4
Bandersnatch said:
Hi Dillion.

Let's take it step by step.

For an object to move in circles (e.g. an orbit), what magnitude must the net force acting on it be?

I hate to say this but I don't even know what you're asking...
 
  • #5
PeroK said:
You have the two key equations. Do you understand them? What does ##F=mv^2/r## mean?

Is it the force something creates while going around in a circle? When I see this equation I think about tension force and a ball on a string being swung around.

F = Gm1m2/r^2 is the gravitational force of an object a certain distance away from the moon.

I think a light bulb just clicked...

G = 6.67 x 10 ^-11
M1 = I will make it up to be a mass of an object let's say 1 kg
M2 = mass of moon which is 7.35 × 10^22 kg
r = 1736482 meters

When I plug those value into F = Gm1m2/r^2, I get F = 2823208.072. I can set this as the F in F = mv^2/r.

2823208.072 = (1kg) (v^2) / 1736482

solving for v^2, v = 2214147.7 m/s

Velocity = distance / time... so 2214147.7 = 10910638.19 / time.

time = 4.9 seconds. is this right?
 
  • #6
You've got the procedure figured out right, but:
Dillion said:
F = Gm1m2/r^2, I get F = 2823208.072.
You've made a mistake in calculations here. Try again, and make sure you plug in everything into the equation. See what result you get then.

Once you've got that done, we can talk about what the equation means, and why it was even a good idea to use it here.
 
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  • #7
Bandersnatch said:
You've got the procedure figured out right, but:

You've made a mistake in calculations here. Try again, and make sure you plug in everything into the equation. See what result you get then.

Once you've got that done, we can talk about what the equation means, and why it was even a good idea to use it here.
Okay I forgot the square the radius :(

but after I redid my calculations I got 1.63 for F

1.63 = (1kg) (v^2) / 1736482

solving for v^2, v = 1682.4 m/s

Velocity = distance / time... so 1682.4 = 10910638.19 / time.

time = 6485.16 seconds. second time the charm?
 
  • #8
Looks about right. How many hours is that?About the equations used: a general fact of nature is that every body moving in circles, no matter what or where it is, is under the influence of a force (or a net result of many forces added together), whose magnitude must be exactly equal to the centripetal force: ##F=mV^2/r##.

The important bit about the centripetal force is that it's just a statement about how strong some other, actual force is. When dealing with circular motion you must find what force is acting on the object (gravity? tension in a string? floor of a space station?), and then you can say with confidence that this force is equal to the centripetal force.

Here, you knew that the object was moving in circles (orbit near the moon surface, which is round), you knew what force was acting (gravity), so all that remained was to say that the only force acting is equal to the centripetal force.
 

1. How long does it take a satellite to orbit the moon?

On average, it takes a satellite about 29 days to orbit the moon. This is because the moon's orbital period around Earth is about 29.5 days, so a satellite orbiting the moon will have a similar orbital period.

2. What factors affect the time of satellite orbit around the moon?

The time of satellite orbit around the moon can be affected by several factors, including the altitude of the satellite's orbit, the gravitational pull of the moon, and the shape of the satellite's orbit. Other factors such as atmospheric drag and solar radiation can also have an impact.

3. Can the time of satellite orbit around the moon change?

Yes, the time of satellite orbit around the moon can change. This can happen due to various factors, such as changes in the satellite's altitude or the moon's gravitational pull. In addition, external forces like solar wind can also cause variations in the orbit time.

4. Is the time of satellite orbit around the moon constant?

No, the time of satellite orbit around the moon is not constant. As mentioned before, it can be affected by various factors, leading to changes in the orbital period. In addition, the moon's own orbit around Earth is not a perfect circle, so the satellite's orbit may also vary slightly due to this.

5. How is the time of satellite orbit around the moon calculated?

The time of satellite orbit around the moon is calculated using Kepler's third law, which states that the square of a planet's orbital period is proportional to the cube of its semi-major axis. By knowing the altitude and shape of the satellite's orbit, the time of orbit can be calculated using this law.

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