A satellite starts in a circular orbit of radius r ( = distance from the center
ID: 2008055 • Letter: A
Question
A satellite starts in a circular orbit of radius r ( = distance from the center of the Earth). Its speed is vj. The gravitational force on the satellite is central, Fr = - GMm/r2 where M is the mass of the Earth; the potential energy is -GMm/r. Suddenly, a burst of the rocket engine increases the speed of the satellite from v1 to v1 + 0.3 v1. Thereafter the satellite moves on an elliptical orbit, as shown. Calculate the apogee ra (= greatest distance from the Earth). Express the answer in terms of the initial radius r. (Show your work please)Explanation / Answer
First start with the circular motion... we know that
m v12 / r = G M m / r2 or v12 = GM/r
Angular momentum must be conserved, so
distance at perigee * speed at perigee = distance at apogee * speed at apogee
r * 1.3 v1 = ra * va
or we can write:
va = 1.3 r v1 / ra square this and we get
va2 = 1.69 r2 (GM/r) / ra2 = 1.69 GM r / ra2
Also, energy must be conserved, so
K + U at perigee = K + U at apogee
½ m (1.3v1)2 - G M m / r = ½ m va2 - G M m / ra
eliminate m everywhere and multiply everything by 2
1.69 v12 - 2GM/r = va2 - 2GM/ra
Sub for v12 and va2
1.69 GM/r - 2GM/r = 1.69 GM r / ra2 - 2GM/ra
divide everything by GM
1.69 /r - 2/r = 1.69 r/ra2 - 2/ra
or
-0.31/r = 1.69 r/ra2 - 2/ra
We are trying to solve for ra so multiply everything by ra2 and divide everything by r
-0.31 ra2 / r2 = 1.69 - 2 ra / r
define x = ra / r and we get
-0.31 x2 = 1.69 - 2 x
or
0 = 0.31 x2 - 2x + 1.69
Using the quadratic formula
x = ( 2 +/- sqrt ( 4 - 4 * 0.31*1.69 )) / 2 * 0.31 = 5.452 or 1.000
We know that x cannot be 1, so x = 5.452 and finally...
ra / r = 5.452 so ra = 5.452 r (final answer!)
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