Calculating Period of Keplerian Orbit on Mystery Planet

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SUMMARY

The discussion focuses on calculating the rotation period of a mystery planet using gravitational data. The planet has a radius of 6502 km and a mass of 4.71 x 1024 kg. An explorer's weight changes from 159 lbs at the equator to 105% of that at the pole, indicating a centripetal force due to the planet's rotation. The correct formula for the period of rotation is derived from centripetal force equations, leading to the conclusion that the period T can be expressed as T = 1/T = √(rΔg/4π2), where Δg represents the difference in gravitational acceleration.

PREREQUISITES
  • Understanding of gravitational force and centripetal acceleration
  • Familiarity with the formula for period of rotation T = 2πR/v
  • Knowledge of basic physics concepts such as mass, weight, and gravitational acceleration
  • Ability to manipulate algebraic equations and solve for variables
NEXT STEPS
  • Research the derivation of centripetal force equations in rotational motion
  • Study the relationship between weight, mass, and gravitational acceleration
  • Explore the application of Kepler's laws in planetary motion
  • Learn about gravitational constants and their role in orbital mechanics
USEFUL FOR

Astronomers, physicists, and students studying planetary motion and gravitational forces will benefit from this discussion, particularly those interested in calculating orbital periods and understanding the dynamics of celestial bodies.

Mirielle
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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:
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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:
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.
 

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