Set 7 Problem number 12


Problem

Problem: Suppose that an Earth satellite of mass 3500 kg, originally in a circular orbit of radius 7.45 * 10 ^ 6 m increases its orbital radius to 7.58 * 10 ^ 6 m. What is the change in its kinetic energy? Based on the strength of the gravitational field at the midway orbital radius, how much work is required to increase its potential energy? How does the change in potential energy compare to the change in kinetic energy?

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Solution

Solution: At orbital distance r the gravitational field strength will be (r / rEarth) ^ -2 times as great as at the surface of the Earth (rEarth stands for the radius of the Earth, approximately 6.38 * 10 ^ 6 meters). At the orbital distances 7.45 * 10 ^ 6 meters and 7.58 * 10 ^ 6 meters, we therefore see that the gravitational field strengths are

distance 7.45 * 10 ^ 6 meters = ( 7.45 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters) ^ -2 * 9.8 m/s^2 = 7.187

and

field at distance 7.58 * 10 ^ 6 meters = ( 7.58 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters) ^ -2 * 9.8 m/s^2 = 6.942

At any point the field strength is equal to the centripetal acceleration v^2 / r; we therefore have

grav accel = v^2 / r,

so

v = `sqrt(grav accel * r).

We obtain velocities vel. at radius 7.45 = `sqrt ( 7.187 m/s^2 * 7.45 * 10 ^ 6 meters) = 7.317 * 10^3 m/s,

and vel. at radius 7.58 = `sqrt ( 6.942 m/s^2 * 7.58 * 10 ^ 6 meters) = 7.253 * 10^3 m/s. For the 3500 kg mass of the satellite we easily obtain kinetic energies of 9.369 * 10 ^ 4 MJ (megaJoules) and 9.206 * 10 ^ 4 MJ. The kinetic energy difference is therefore

KE change = ( 9.206 * 10 ^ 4 - 9.369 * 10 ^ 4) MJ = -16 MJ. The midpoint orbital radius is 7.515, which implies a midpoint gravitationalfield of ( 7.515 / rEarth) ^ -2 * 9.8 m/s^2 = 7.063 m/s^2, and agravitational force of 3500 kg * 7.063 m/s^2 = 2.472 * 10 ^ 4 N.

At the orbit is 'raised' the distance parallel to the gravitational field is the difference ( 7.58 - 7.45) * 10 ^ 3 meters = .1300 meters of the orbital radii.

The work required to 'raise' the orbit against the gravitational pull will therefore be 2.472 * 10 ^ 4 N * .1300* 10 ^ 6 meters = 3213 MJ.

The ratio of this potential energy change to the kinetic energy change is thus 3213 MJ / ( -16 MJ) = -2.81.

The ratio computed here is based on a midpoint approximation to an inverse square gravitational field and if computed to a sufficient number of decimal places will differ from the precise ratio, which is exactly -2. The change of KE in a transition of circular orbits is always exactly half the magnitude of the potential energy change, and of the opposite sign. That is, the kinetic energy loss is always exactly half of the kinetic energy gain.

Generalized Solution

Generalized Response: A circular orbit of radius r1 will have orbital velocity v1 such that v1^2 / r1 = g1, where g1 is the acceleration of gravity, or gravitational field strength, at distance r1 from the center of the Earth. Since the ratio of gravitational fields at two distances is the inverse square of the distance ratio, at radius r1 the field is

g1 = (grav. field at distance r1) = (r1 / rEarth) ^ -2 * g,

where g is the gravitational field strength at the surface of the Earth. The squared velocity of the orbit whose radius is r1 is therefore

square of velocity in first orbit = v1^2 = g1 * r1 = (r1 / rEarth) ^ -2 * g * r1 = rEarth^2 / r1 * g

and the kinetic energy is

KE of first orbit = .5 m v1^2 = .5 * m * rEarth^2 / r1 * g.

Similarly the kinetic energy in the second orbit is

KE of second orbit = .5 m v2^2 = .5 * m * rEarth^2 / r2 * g.

The difference in the kinetic energies is

KE difference = KE2 - KE1 = .5 * m * rEarth^2 (1 / r2 - 1 / r1) * g.

The gravitational field midway between orbital radii r1 and r2 is

gMid = (rMid / rEarth) ^ -2 * g,

where rMid = (r1 + r2) / 2. The force on the mass m at this distance is therefore

Fmid = m * gMid = m * (rMid / rEarth) ^ -2 * g,

and the work required to 'raise' the mass through the required distance (r2 - r1) parallel to the gravitational field is

work against gravity = Fmid * (r2 - r1) = m * (rMid / rEarth) ^ -2 * g * (r2 - r1) = rEarth^2 / [(r1 + r2) / 2] * g * (r2 - r1)

PUT AN END TO THIS FOOLISHNESS!!!!

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