Kepler's Law and Non-terrestrial Orbits (Not Earth)

In summary: The period can be calculated using Kepler's third law, which relates the orbital period, the distance between the satellite and the planet's center, and the mass of the planet. The solution provided assumes that the planet in question is the Earth and uses the known values for the Earth's mass and the distance given in the problem to calculate the period in Earth days.
  • #1
Axoren
17
0

Homework Statement


A satellite orbits a planet at a distance of 6.80 multiplied by 10^8 m. Assume that this distance is between the centers of the planet and the satellite and that the mass of the planet is 3.08 multiplied by 10^24 kg. Find the period for the moon's motion around the earth. Express the answers in Earth days

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.

I'm incredibly sure this is a correct value, I've done multiple calculations and always got within 0.1% of this value

I saw this question in the archive, but I don't believe anyone realized the answer was in Earth days or they just did not show their work (which is what I was hoping to see).

Please help.

Found my problem, needed to divide by days represented in seconds (Period / (60 * 60 * 24));

Answer was 90 days. (decimal dropped out of laziness, forgive me significant digits.)
 
Last edited:
Physics news on Phys.org
  • #2
HOW did you get it "in days of this planet" when nothing is said about the planet's rotation?
 
  • #3
Axoren said:

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.
How did you go from Kepler's third law to 7779136.049? What are the units for this number?

In other words, show your work. It's a bit difficult to find where your mistake is if you don't show us how you arrived at your result.
 
  • #4
Axoren said:

Homework Statement


A satellite orbits a planet at a distance of 6.80 multiplied by 10^8 m. Assume that this distance is between the centers of the planet and the satellite and that the mass of the planet is 3.08 multiplied by 10^24 kg. Find the period for the moon's motion around the earth. Express the answers in Earth days

Homework Equations


Kepler's 3rd Law
Period² = (4(pi)radius³)/(Gravitational Constant * Mass of Massive Body)

The Attempt at a Solution



I have the answer in days of this planet, however, I don't know how to get the answer in Earth days.

7779136.049 = Period.
How did you get the radius of planet X if the question on,y mentions the satellite's orbital distance?
 
  • #5
Mrrss24 said:
How did you get the radius of planet X if the question on,y mentions the satellite's orbital distance?

That's not the planet's radius, it's the satellite's orbital radius.
 

1. What are Kepler's Laws?

Kepler's Laws are a set of three laws that describe the motion of planets and other objects in the solar system. They were formulated by German astronomer Johannes Kepler in the early 17th century.

2. What is the first law of Kepler?

The first law, also known as the law of orbits, states that all planets move in elliptical orbits with the sun at one focus.

3. What is the second law of Kepler?

The second law, also known as the law of areas, states that a line connecting a planet to the sun will sweep out equal areas in equal times, meaning that a planet moves faster when it is closer to the sun and slower when it is farther away.

4. What is the third law of Kepler?

The third law, also known as the law of harmonies, states that the square of a planet's orbital period is directly proportional to the cube of its average distance from the sun. This means that planets farther from the sun have longer orbital periods.

5. Do Kepler's Laws apply to non-terrestrial orbits?

Yes, Kepler's Laws apply to all objects in the solar system, regardless of whether they are planets or not. They also apply to objects in other solar systems and even galaxies, as long as they are orbiting a central body.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
972
  • Classical Physics
Replies
2
Views
803
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
995
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
445
  • Introductory Physics Homework Help
Replies
16
Views
2K
Back
Top