Spacecraft in circular orbit

[Kelseywelsey]

Homework Statement

A spacecraft of mass 1000 kg is in an elliptical orbit about the earth. At point A the spacecraft is at a distance of 1.2 x 10^7 meters form the center of the Earth and its velocity ( 7.1 x 10^3 m/s) is perpendicular to the line connecting the center of the Earth to the spacecraft . Mass of the earth= 6.0 x 10^24 and radius= 6.4 x 10^4. I need to find the speed of the spacecraft if it is in a CIRCULAR orbit and the minimum speed of the craft at pt. A if it is to escape completely from the earth.

Homework Equations

V at point B= 2400 m/s

The Attempt at a Solution

So far, I've figured that the mechanical energy is 8.1 x 10^9 J
Angular momentum= 8.52 x 10^13
and speed at a point directly across from A
V= 2.4 x 10^3

 

[Mindscrape]

Alright, so more than just telling us what numbers you have, you should go over some concepts and why you chose these numbers.

What kind of questions do you have that could be answered and potentially helpful to you solving the problem?

More importantly, what is it that is keeping the spacecraft in orbit? How do you think that case will be broken?

 

[m2003]

m/s

First, it is important to clarify the question. Is the spacecraft currently in a circular orbit or an elliptical orbit? In the statement, it says the spacecraft is in an elliptical orbit, but in the homework equations, it mentions the speed at a point directly across from point A, which would only be relevant if the spacecraft is in a circular orbit. For the sake of this response, I will assume that the spacecraft is currently in an elliptical orbit, as stated in the problem.

To find the speed of the spacecraft in a circular orbit, we can use the formula for circular orbital velocity: v = √(GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth to the spacecraft. Plugging in the values given in the problem, we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2)(6.0 x 10^24 kg)/(6.4 x 10^6 m + 1.2 x 10^7 m)]

v = 7.37 x 10^3 m/s

Therefore, the speed of the spacecraft in a circular orbit is 7.37 x 10^3 m/s.

To find the minimum speed of the spacecraft at point A in order to escape completely from the Earth, we can use the formula for escape velocity: v = √(2GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth to the spacecraft. Plugging in the values given in the problem, we get:

v = √[(2)(6.67 x 10^-11 Nm^2/kg^2)(6.0 x 10^24 kg)/(6.4 x 10^6 m + 1.2 x 10^7 m)]

v = 1.11 x 10^4 m/s

Therefore, the minimum speed of the spacecraft at point A to escape completely from the Earth is 1.11 x 10^4 m/s. This means that if the spacecraft's speed at point A is greater than 1.11 x 10^4 m/s, it will escape from the Earth's gravitational pull and enter into an orbit around the sun.