Calculating Period of Keplerian Orbit on Mystery Planet

  • Thread starter Mirielle
  • Start date
  • Tags
    Orbits
In summary, an explorer on a planet with a different gravity (1.59 hr) experiences a different weight depending on where he is on the planet.
  • #1
Mirielle
1
0
Hi, I have a question on calculating the period of a planet where I know the gravity at the equator and at the pole.

Here is the known information for the planet:
Planet radius: [itex]6502 km[/itex]
Planet mass: [itex]4.71 \ast 10^{24} kg[/itex]

An explorer, who weighs 191 lbs on earth, stands at the equator of this mystery planet. He finds that he currently weighs 159 lbs.
He then moves to the north pole of the planet and finds that he weighs 105% of what he previously weighed.
The question then asks what the period of rotation for the planet is.

So, I thought the following formula would probably be useful in solving this.
[tex]v = \frac{2 \pi R}{T}[/tex]
Or in this case:
[tex]T = \frac{2 \pi R}{v}[/tex]
So, I already know the radius, which only leaves figuring out the velocity with which it is spinning...[tex]a_{equator}= \frac{Gm}{r^2}[/tex]

[tex]a_{pole}= 1.05 a_{equator}[/tex]

We also know that:
[tex]a_{centrip.}= \frac{v^2}{r}[/tex]

So, now I solve for v by combining these...

[tex]\frac{1.05 GM }{r^2 }= \frac{v^2 }{r }[/tex]

which gives me:

[tex]v = \sqrt{\frac{1.05 GM}{r}}[/tex]

Now I plug in the following values from above:

[tex]r = 6502 km = 6502000 m[/tex]

[tex]m = 4.71 \ast 10^{24} kg[/tex]

(And the constant)

[tex]G = 6.673 \ast 10^{-11}[/tex]

And I find that v = 7125 m/s

Now, I convert this to hours due to the period being in hours.
v = 7125 m/s = 25650000 m / hr

Then we return to this equation I presented before:

[tex]T = \frac{2 \pi R}{v}[/tex]

And plugging in the values...

[tex]T = \frac{2 \pi (6502000)}{25650000} = 1.59 hr[/tex]

However, this is wrong.
I'm curious to know where I'm going wrong here.

P.S. this is the first time I've ever used TeX, I hope it works properly!
 
Last edited:
Physics news on Phys.org
  • #2
Mirielle said:
Here is the known information for the planet:
Planet radius: [itex]6502 km[/itex]
Planet mass: [itex]4.71 \ast 10^{24} kg[/itex]

An explorer, who weighs 191 lbs on earth, stands at the equator of this mystery planet. He finds that he currently weighs 159 lbs.
He then moves to the north pole of the planet and finds that he weighs 105% of what he previously weighed.
The question then asks what the period of rotation for the planet is.
Suggestion: Set out the equation for T in algebraic form and then plug in numbers.

In this case, the difference between the weight at the equator and at the pole is the centripetal force due to the rotation of the planet. So you don't really have to worry about what the force of gravity is.

[tex]m\Delta g = F_c = m\omega^2r = m\frac{4\pi^2}{T^2r}[/tex]

We know what his weight is on Earth and g for the earth, so we can determine his mass. From that you can determine the value for g on the equator and at the pole

[tex]\Delta g = \frac{4\pi^2}{T^2r}[/tex]

[tex]1/T = \sqrt{r\Delta g/4\pi^2}[/tex]

AM
 
Last edited:
  • #3
The weight at that point is the sum of two forces. You tried to use centripetal force to explain the entire weight. I'll leave it up to you to fill in the rest.
 

Related to Calculating Period of Keplerian Orbit on Mystery Planet

1. How do you calculate the period of a Keplerian orbit?

To calculate the period of a Keplerian orbit, you need to know the mass of the central body, the distance between the two bodies, and the gravitational constant. Using these values, you can use Kepler's third law, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis. This can be expressed in the equation T^2 = (4π^2a^3)/GM, where T is the period, a is the semi-major axis, G is the gravitational constant, and M is the mass of the central body.

2. What is a Keplerian orbit?

A Keplerian orbit is an elliptical orbit around a central body, where the orbiting body follows Kepler's laws of planetary motion. These laws were developed by Johannes Kepler in the 17th century and describe the motion of planets and other celestial bodies in our solar system.

3. How is the period of a Keplerian orbit affected by the distance between the two bodies?

The period of a Keplerian orbit is directly proportional to the distance between the two bodies. This means that as the distance between the two bodies increases, the period of the orbit also increases. This relationship is described by Kepler's third law, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis.

4. Can the period of a Keplerian orbit change over time?

Yes, the period of a Keplerian orbit can change over time. This can happen due to various factors, such as the presence of other celestial bodies, gravitational perturbations, and changes in the mass or distance of the central body. However, these changes are usually small and can be calculated and predicted using mathematical equations.

5. Is the period of a Keplerian orbit the same for all planets?

No, the period of a Keplerian orbit is not the same for all planets. It varies depending on the mass and distance of the central body. For example, the period of Earth's orbit around the Sun is approximately 365 days, while the period of Mars' orbit is approximately 687 days. However, all planets in our solar system follow Kepler's laws of planetary motion and have elliptical orbits around the Sun.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
843
  • Introductory Physics Homework Help
Replies
6
Views
707
  • Introductory Physics Homework Help
Replies
1
Views
852
  • Introductory Physics Homework Help
Replies
4
Views
611
  • Introductory Physics Homework Help
Replies
17
Views
431
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top