Mercury rotating the earth in an empty region of the galaxy

In summary, the conversation discusses the hypothetical scenario of Earth and Mercury being pulled out of the solar system and orbiting each other in an empty region of the galaxy. The question then arises about Mercury's orbital period if its distance from Earth remains constant and their masses are unchanged. In addition, there is a question about the kinetic energy of this orbiting system and how it compares to orbits predicted by Kepler's First Law. The equations used to solve these questions are also mentioned. The estimated period for Mercury's orbit is 110 days, and the calculated kinetic energy is 6.77 * 10^28 J. There is uncertainty about the units used and how to compare the circular motion used in the equations to the elliptical motion described in
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Verdict
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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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  • #2
I also have a completely different question, should I just make a separate thread for that?
 

1. What is the distance between Mercury and Earth?

The distance between Mercury and Earth varies depending on their positions in their respective orbits. On average, Mercury is about 57.9 million kilometers away from Earth. However, at their closest approach, they can be as close as 77.3 million kilometers and at their farthest, they can be 222.1 million kilometers apart.

2. How does Mercury's rotation affect Earth?

Mercury's rotation does not directly affect Earth as they are both independent celestial bodies. However, their gravitational pull on each other can cause slight perturbations in their orbits.

3. Is Mercury the only planet that rotates around Earth in an empty region of the galaxy?

No, Mercury is not the only planet that rotates around Earth in an empty region of the galaxy. All the planets in our solar system, including Earth, orbit around the sun, which is located in a relatively empty region of our galaxy.

4. How long does it take for Mercury to rotate around Earth?

Mercury takes about 88 days to complete one orbit around the sun. As it orbits, it also rotates on its axis, which takes about 59 Earth days to complete. Therefore, it takes about 59 Earth days for Mercury to complete one rotation around Earth.

5. Can Mercury ever collide with Earth while rotating in the same region of the galaxy?

No, Mercury and Earth are not on a collision course. As mentioned earlier, their orbits are independent of each other and their distances from each other vary greatly. There is no known scenario where they would collide while rotating in the same region of the galaxy.

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