A spaceship of mass 10^6 kg is in circular orbit 10^6 m from...

Uploaded: 2022-10-17T16:57:18-08:00
Duration: 5 min 36 s
Channel: Krystal K

A spaceship of mass $10^6$ kg is in circular orbit $10^6$ m from the center of a planet of mass $6 \times 10^{25}$ kg. A shuttlecraft of mass $4 \times 10^4$ kg is resting on the planetary surface. The planet has a radius of $2 \times 10^5$ m. (a) Calculate the total mechanical energy of the spaceship in orbit about the planet -- given that its orbital velocity is $6.4 \times 10^4$ m/sec. (b) How much energy would have to be imparted to the shuttlecraft sitting on the planet's surface (presumably by its own engines) to put it into the same orbit as the spacecraft? (Hint: You must first calculate the total energy which the shuttle needs to attain the circular orbit.)

Transcript

00:01 Okay, so we have a spaceship in orbit around earth at a height of 220 kilometers, and then it's going to move to a height of 370 kilometers above the surface.

00:11 And we want to know what the work is.

00:15 And that's just the change in energy that it would need in order to make that trip to that higher distance.

00:29 So they give us the mass of the spaceship, which is 3 ,500 kilograms, the mass of the earth, and then i also have the radius of the earth over here, because we'll need that.

00:39 So the work for this problem is going to be the change in the energy.

00:48 And so obviously when you have a change in something, it's going to be e2 minus e1.

00:55 And for this problem, the energy of something orbiting the earth is going to be negative g times the mass of the spaceship times the mass of earth over the distance from the center of the earth to, so that's the radius of earth, to wherever the spaceship is.

01:28 So in order to get that total distance, you have to have the radius plus the height above the surface.

01:35 And that is multiplied by two.

01:39 All right.

01:39 So we'll have to find the energy at each of these heights and just subtract one from the other to find that difference.

01:49 So the e1 is going to be g, which is 6 .67 times 10 to the negative 11.

02:06 Newtons times meter squared over kilograms squared times the mass of the spaceship, which is 3 ,500 kilograms times the mass of the earth, 5 .97 times 10 to the 24 kilograms.

02:29 And that is going to be divided by two times, so the radius of the earth, which is 6 .37 .7...

Question

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