Mercury rotating the earth in an empty region of the galaxy

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SUMMARY

The discussion centers on the hypothetical scenario of Earth and Mercury orbiting each other in an empty region of the galaxy, specifically analyzing Mercury's orbital period and kinetic energy. Using the formula T = 2∏ * (a^3 / μ)^0.5, the calculated orbital period is approximately 9,536,755.185 seconds, or 110 days. The kinetic energy of the system is derived from K = 1/2 * I * ω², yielding a value of approximately 6.77 * 10^28 J. The comparison to Kepler's First Law highlights the differences between circular and elliptical orbits.

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A: Homework Statement

If the Earth and Mercury happened to be pulled out of the solar system by the gravitational force of a passing neutron star, and then left in orbit about each other in an otherwise empty region of the galaxy, our lovely planet would probably get very cold. But assuming that someone out there was still interested in such things, what would be Mercury's orbital period if its distance from the Earth was a constant value of 9.72 * 10^8 meters and their respective masses are unchanged? (Using that the mass of the Earth is 5.9736 * 10^24 kg and the mass of mercury is 3.3022 * 10^23 kg)

B: What is the kinetic energy of this orbiting Mercury-Earth system in the rest frame of their center mass?

C: How does this orbit compare to orbits predicted by Kepler's First Law if Mercury is assumed to be the satellite and the Earth to be a 'sun'?

Homework Equations


A: My thoughts were that for this one, you could use the equation
T= 2∏ * (a^3 / μ)^0,5
Where T is the period in seconds, a is the distance between the Earth and mercury and μ
is the gravitational constant multiplied by the mass of the earth.

B: K = 1/2 * I * ω²
Where K = kinetic energy, I = moment of inertia and ω = angular speed
From here on I'm not entirely sure, but I would use the following:
I = Mmerc * R² where Mmerc is the mass of mercury and R is the distance between Mercury and the Earth, given in the question.
ω = 2 ∏ / T, where T is the period in seconds that we found in question A.

The Attempt at a Solution


A: Filling in all the information gives me a period of 9536755,185 seconds, converted to days this would be 110 days.

B: Filling in all the information gives me that I = 3.119866 * 10^41 kg*m² (not sure about the units), ω = 6.588389 * 10^-7
So that K would be 6,771180 * 10^28 J, or in the right number of significant figures 6,77 * 10^28 J, which sounds like a lot to me, so this is quite probably wrong. Could anyone give me some advice here?

C: Honestly, I don't really know what to say here. I know that I used a circular motion for these questions instead of an elliptical one, so that means that both foci are located at the same point, the circles center, and that the major and minor radii are therefore the same. Kepler's first law uses an ellipse where the sun is located on one of the focii, and the planet would be located on the actual ellipse. How should I compare these orbits?
 
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I also have a completely different question, should I just make a separate thread for that?
 

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