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Phy 121
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Question regarding mass & KE
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Problem
What are the KE and PE changes and the ratio of PE to KE change if an Earth satellite of mass `mass kg, originally in a circular orbit of radius 7.98 * 10 ^ 6 m, increases its orbital radius to 8.17 * 10 ^ 6 m?
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 7.98 * 10 ^ 6 meters and 8.17 * 10 ^ 6 meters, we therefore see that the gravitational field strengths are
distance 7.98 * 10 ^ 6 meters = ( 7.98 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters)^ -2 * 9.8 m/s^2 = `g1
field at distance 8.17 * 10 ^ 6 meters = ( 8.17 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters) ^ -2 * 9.8 m/s^2 = 5.976.
At any point the field strength is equal to the centripetalacceleration v^2 / r; we therefore have
grav accel = v^2 / r, which solved for v gives us
v = `sqrt(grav accel * r).
We obtain velocities
vel. at radius 7.98 = `sqrt ( 6.264 m/s^2 * 7.98 * 10 ^ 6 meters) = 7.07 * 10^3 m/s, and
vel. at radius 8.17 = `sqrt ( 5.976 m/s^2 * 8.17 * 10 ^ 6 meters) = 6.987 * 10^3 m/s
For the `mass kg mass of the satellite we easily obtain kinetic energies of 87470 MJ (megaJoules) and 85430 MJ. The kinetic energy difference is therefore
KE change = ( 85430 - 87470) MJ = -2040 MJ
The midpoint orbital radius is 8.074, which implies a midpoint gravitational field of
midpoint field = ( 8.074 / rEarth) ^ -2 * 9.8 m/s^2 = 6.119 m/s^2,
and gravitational force of
midpoint gravitational force = `mass kg * 6.119 m/s^2 = 21410 N.
At the orbit is 'raised', the distance parallel to the gravitational field is the difference of the orbital radii:
distance parallel to field = ( 8.17 - 7.98) * 10 ^ 3 meters = .19 meters
The approximate work required to 'raise' the orbit against the gravitational pull will therefore be
approximate work to 'raise' orbit = 21410 N * `dr* 10 ^ 6 meters = 4067 MJ
Since the PE change is the work done by the satellite against gravity, this is the potential energy change of the satellite.
The ratio of this potential energy change to the kinetic energy change is thus
`dPE / `dKE = 4067 MJ / (-2040 MJ) = -1.994.
The ratio computed here is based on a midpoint approximation to an inverse square gravitational field. If the ratio is based on a more accurate approximation (e.g., on a subdivision of the distance into a sufficient number of subintervals, with a midpoint approximation on each), the approximate ratio will approach the precise ratio which is exactly -2.
University Physics students: This ratio is computed using
KE = .5 m v^2 with the velocity v = `sqrt(G M / r) obtained from setting centripetal acceleration equal to gravitational attraction, and
`dPE = work required to change orbital distance, which is the integral of force * distance
`dPE = int( G M m / r^2, r, r1, r2) = G M m (1 / r1 - 1 / r2).
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
The KE of an orbit is .5 m * G M / r, which can be obtained from v = `sqrt( G M / r), obtained by setting centripetal force equal to gravitational force; however to illustrate the use of proportionality we use a different and perhaps more cumbersome approach. This approach will express the orbital velocity and KE in terms of g = 9.8 m/s^2 and the ratio rEarth / r of Earth radius to orbital radius. These end results involve easily visualized quantities are in many ways more intuitive than the equation .5 m * G M / r.
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 g * [ rEarth^2 / r1].
Similarly the kinetic energy in the second orbit is
KE of second orbit = .5 m v2^2 = .5 * m g * [ rEarth^2 / r2].
The difference in the kinetic energies is
KE difference = KE2 - KE1 = .5 * m g * rEarth^2 (1 / r2 - 1 / r1) .
To find the approximate PE difference we use the midpoint gravitational field (the field at r halfway between r1 and r2) to estimate the average force required to 'raise' the satellite, then multiply approximate force by distance.
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.
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) = m g * (rEarth / rMid) ^ 2 * (r2 - r1)
Since rMid between r = r1 and r = r2 is (r1 + r2) / 2 we have
`dPE = work against gravity = m g * rEarth^2 / [(r1 + r2) / 2] ^ 2 * (r2 - r1).
It can be shown that if r2 / r1 is nearly 1, the expression (r2 - r1) / [ (r2 + r1) / 2 ] ^2 is very nearly equal to 1 / r1 - 1 / r2, and we have
`dPE = m g * rEarth^2 ( 1 / r1 - 1 / r2).
The techniques of Calculus (integrating force * distance from r1 to r2, with force = m g * ( rEarth / r ) ^ 2 ) give the same result for any r1 and r2.
More generally for any planet of mass M and satellite of mass m we have `dPE = G M m ( 1 / r1 - 1 / r2 ) and `dKE = - .5 G M m ( 1 / r1 - 1 / r2 ).
Comparing the expressions for `dPE and `dKE, we see that `dPE / `dKE = -2. That is,
between two circular orbits the potential energy change has exactly twice the magnitude of the kinetic energy change, and the opposite sign.
For increasing orbital distance PE increases by a certain amount and KE decreases by half that amount.
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This is the problem. I have taken the answers for the KE for each velocity and solved for the mass. That mass is then consistent the other KE and the midpoint gravitation force, but I am not sure how the mass is obtained.
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I have gotten different masses depending on when I click on the problem set question, I am sure I am just missing the obvious here. I understood your answer to this question earlier for the lab as to just put in 1 kg for mass? Is that right? I know it will not work on this question.
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The problem randomizer appears not to be generating a mass.
However if we have
vel. at radius 7.98 = `sqrt ( 6.264 m/s^2 * 7.98 * 10 ^ 6 meters) = 7.07 * 10^3 m/s, and
vel. at radius 8.17 = `sqrt ( 5.976 m/s^2 * 8.17 * 10 ^ 6 meters) = 6.987 * 10^3 m/s
and if for the `mass kg mass of the satellite we 0btain kinetic energies of 87470 MJ (megaJoules) and 85430 MJ
then since the kinetic energy difference is
1/2 m v2^1 - 1/2 m v1^2 = 1/2 m ( v2^2 - v1^2)
we have
1/2 m (v2_2 - v1^2) = ( 85430 - 87470) MJ = -2040 MJ
so that
m = 2 * (-2040 MJ ) / ( (6.987 * 10^3 m/s)^2 - (7.07 * 10^3 m/s)^2 ),
which is easily found by evaluating the expression (remember the MJ is 10^6 J).
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