newtons version on keplers third law

Question

Newton’s version of Kepler’s third law is: P² = 4 π² / [G (M₁ + M₂)] × a³. The space shuttle orbits 271 km above the Earth's surface. How often do the astronauts see a sunrise (in minutes)?

Use the gravitational constant G = 6.67 × 10^-11 m³ kg^-1 s^-2, the mass of the Earth M = 5.97 × 10²⁴ kg, and the radius of the Earth to be 7000 km.

Answer 1

It seems you asked this a bit ago, but in case you are still looking for an answer or don't understand, I will at least provide a path to the answer.

You have all of the information you need, it's just a bit sneaky on how and why. You're trying to figure out P, which leaves us with needing G, M₁, M₂, and a.

1) What variables do we already know?

G is already given, and we can plug in the mass of the Earth (M), as either M₁ or M₂, I'll simply choose M₁.

2) What variables do we need?

It looks like we still need M₂ and a. Let's start with a.

The way this question is worded implies that the orbit is perfectly circular, since they are only giving you one radius. The shuttle orbits 271 km ABOVE the surface of the Earth, and the radius of the earth is given to be 7000 km, so we simply add those two together to get a, as that is the orbital distance for the space shuttle to the center of the Earth.

But what about M₂? Recall, even though we are talking about Newton's laws, one of Kepler's laws proves that mass does NOT matter when it comes to the orbital period of an object (at least, when it's relatively very VERY tiny). It could be a baseball or a cruise ship: if you put multiple objects of different mass into orbit, as long as they were orbiting at the same radius, their velocities would all be the same, and by extension, their orbital period as well.

So in short, M₂ does not matter (no pun intended), and can be considered to be 0. With that, you should have enough information to solve the equation. Note: you would only throw in M₂ if the second mass was at least somewhat comparable in size, like Pluto and it's moons, or even the Earth and our Moon (although the difference is tiny. The dominating variable that REALLY changes things up is going to be a, since it's to the power of 3)

Just to emphasize, let's say you WERE given the mass of the shuttle (about 2 million kg), that mass is 3.35x10^-19 times smaller than the mass of the Earth. Not even NASA will use decimals past 15 places, as it's not considered significant at that point, and that number is 19 decimals!