Changing orbit of a space shuttle

Question

In the figure here, a space shuttle is initially in a circular orbit of radius r about Earth. At point P, the pilot briefly fires a forward-pointing thruster to decrease the shuttle’s kinetic energy K and mechanical energy E. (a) Which of the dashed elliptical orbits shown in the figure will the shuttle then take? (b) Is the orbital period T of the shuttle (the time to return to P) then greater than, less than, or the same as in the circular orbit?

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Hello. I have a major trouble understanding this question. How do we know that the space shuttle passes throught the point P after losing speed ? Also, the lower orbits must have greater speeds. Doesn't that mean the pilot must increase the shuttle's kinetic energy to go to lower orbits ? Appreciate any help! Thanks!

Answer 1

It is assumed that the spacecraft fires changes its velocity in an instant, not over a period of time. Its velocity is decreased exactly at the point in time it passes through P.

It is true that spaceships in lower circular orbits have greater orbital velocities, but in elliptical orbits the velocity changes with the distance between the two masses (since the potential energy increases with distance, the kinetic energy must decrease).

It is therefore the case that a spacecraft in the orbit 1 has a lower velocity at point P than a spacecraft in the initial Orbit at point P.(Think about it, they are both the same distance away, so the potential energy is the same)

Answer 2

You're on the right track.

"How do we know that the space shuttle passes throught the point P after losing speed ?"

The assumption (in these types of problems) is that the thruster is applied for a very short time compared to the duration of the orbit. In that way, we can assume that applying the thruster is effectively instantaneous. So if the shuttle applies the thruster at point P, we can assume that once they're done, they will still be (roughly) at point P. Then, because of the nature of the two-body problem --- they are assured to come back to the same point P.

So what happens to the kinetic energy before and after the thruster fires (while still at point P)?
What happens to the potential energy?
The total energy?

What does that tell you about the resulting orbit?

Answer 3

All you seem to have never ever been able to see the absurdity of the question when it says that the pilot briefly fires a forward-pointing thruster to decrease the shuttle's kinetic energy " AND MECHANICAL ENERGY" while Mechanical energy is conserved for conservative forces , in this case is the gravitational force

Firing front-facing thruster means decreasing speed , firing back-facing thruster means increasing speed, firing side-facing thruster means to change the direction Here in this case, it is to dcrease the spcaceship's kinetic energy which means to decrease its speed by firing forward-facing thruster Increasing and decreasing in orbits around the Earth is all about changing the distance between the Earth's centre of mass and the orbiting object The total energy or the mechanical energy is conserved or constant , no matter how near or far away from the Earth's centre of mass the orbiting object might be , within the Earth's gravitational field. It means that E(mec) at a lower altitude and E(mec) at a higher altitude are the same. Only kinetic and gravitational potential energy change, but their sum which is mechanical energy E(mec) is unchanged

delta E = Ef -Ei = 0

Where Ef is the total energy of the object at the final position, and Ei is the total energy at the initial position

E(mec) = KE + Ug . So the Kinetic energy KE at point P in the initial circular orbit with the distance " R1 " from tge Earth's centre is K = GMm/2R1 And the speed can be found from the equation F(R) = ma(R).
Where F(R) is gravitaional force on the spaceship , a(R) is radial acceleration of the spaceship toward the Earth's cent5 as it travels in circular orbit around the Earth F(R) = ma(R) F(R) = GMm/R^2, and a(R) = v^2/R GMm/R^2 = mv^2/R---->GM/R = v^2 -----> v = (MG/R)^1/2

Whenever kinetic energy decreases, it means that the speed v decreases And the only way for the speed of the spaceship to decrease is that the distance R between the Earth's centre and the spaceship must increase which means that the spaceship will have to change its orbit to a higher orbit for the distance between the Earth's centre and itself to be farther than its initial circular orbit, but not to any lower orbit where its speed will certainly increase as the distance between the Earth's centre and itself decreases So after firing the front-facing thruster at point P, the spaceship changes its original circular orbit to the larger elliptical orbit where it speed slows down , its kinetic energy decreases and its gravitational potential energy increases as the distance R between itself and the Earth's centre increases Looking at this equation of the spaceship's total or mechanical energy can help you see more clearly how why it has to change to a higher orbit, but not a lower orbit after firing tge front thruster

Emec= KE + Ug = GMm/2R - GMm/R

Whenever R increases---> KE= GMm/2R decreases, and Ug = - GMm/R increases because smaller negative value of Ug means the value of gravitational potential energy Ug increases and that is why the total energy "Emec" remains the same

You can replace R with " a " as semimajor axis for elliptical orbits, but it won't make any difference