A couple of gravitational problems (kelper's third law)

In summary, the first conversation discusses the calculation of a satellite's period of motion when placed in orbit around the Earth at an altitude of 1.6 x 10^6m, using Kepler's third law. The calculated period is 7.2 x 10^3s. The second conversation involves determining the radius of a satellite's synchronous orbit around Uranus, using the same law and given the average radius and period of revolution of Miranda, one of Uranus' moons. The calculated radius is 5.8 x 10^7m. However, there are some discrepancies in the numbers used, as the orbital periods and radii should refer to objects orbiting the same gravitational center.
  • #1
Vickyvicksss
5
0

Homework Statement



A satellite is to be placed in orbit about the Earth at an altitude of 1.6 x 10 6m. What will be its period of motion?

The answer is 7.2x 103s

2. The attempt at a solution
Using kelper's third law.

R3 / T2 = R3 / T2

Radius of satellite: 6.38 x 106 + 1.6 x 106 = 7.98 x 106
Radius of earth: 6.38 x 106
Period of satellite: ?
Period of earth: 3.16 x 107

When I plugged these numbers in, I get 4.42 x 107s
I've used the formula before without any problems, so that's not the issue.


3. Homework Statement
Determine the radius of a satellite's synchronous orbit around Uranus. Your very friendly, local astronomer informs you that uranus' moon, Miranda, has an average radius of revolution of 1.24 x 105 km with a period of revolution of 1.4 Earth days.

The answer is 5.8 x 107m

4. The attempt at a solution
Same thing as above, I used kelper's third law.
R3 / T2 = R3 / T2

Radius of Miranda: 1.24 x 105
Radius of satellite: ?
Period of miranda: 1.4 days * 3600 * 24 = 120960s
Period of satellite: (I'm assuming this is period of Uranus?) 2.66 x 109

I plugged these numbers in and I got 9.73 x 107m



Thank you in advance!~~ :)
 
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  • #2
Vickyvicksss said:
Period of earth: 3.16 x 107
What do you mean by 'period of earth'? Where did you get this number?
Period of satellite: (I'm assuming this is period of Uranus?) 2.66 x 109
Where did you get this number?
 
  • #3
The correct number to plug into Kepler's law here, given that you have plugged in Earth's radius for one of the Rs, is the orbital period of a satellite at ground level, not the orbital period of the Earth around the Sun. The numbers always must refer to orbital periods and orbital radii around the same gravitational center - the Earth, in this case.

Apart from the satellite, what goes round the Earth that has an orbital radius and period that you could look up?

For the second question, do you know what a synchronous orbit is?
 

Related to A couple of gravitational problems (kelper's third law)

What is Kepler's third law?

Kepler's third law, also known as the harmonic law, states that the square of a planet's orbital period is directly proportional to the cube of its semi-major axis. This means that the farther a planet is from the sun, the longer it takes to complete one orbit.

How did Kepler come up with this law?

Johannes Kepler, a German astronomer and mathematician, discovered this law while studying the motion of planets around the sun. He analyzed and compared the orbital periods and distances of different planets to come up with this law.

What are some real-life applications of Kepler's third law?

Kepler's third law is used to calculate the orbital periods and distances of planets, moons, and other celestial bodies in our solar system. It is also used in space missions to plan and predict the trajectories of spacecrafts.

Does Kepler's third law only apply to our solar system?

No, Kepler's third law applies to any system where two objects are orbiting each other under the influence of gravitational force. This law has been successfully applied to other stars and their planets, as well as binary star systems.

What are some limitations of Kepler's third law?

Kepler's third law assumes that the objects are in circular or elliptical orbits and that there are no external forces acting on them. In reality, the orbits of planets and other celestial bodies may be affected by other factors such as gravitational pull from other objects, making the law less accurate.

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