|
HINTS FOR PROBLEM SET #9
Problem Set #9 – Problem 1 Since the only force acting on the probe is a gravitational force and this is a conservative force, you can use the Work Energy Theorem in this form: TA + VA = TB + VB This is sometimes called the Conservation of (Mechanical) Energy. TA = ½ mvA2 VA = (- G m Mearth) / rA TB = ½ mvB2 VB = (- G m Mearth) / rB Note: vA, rA and rB are known. Rearth = (3960 miles) x (5280 ft/mile) = 20,908,800 ft G Mearth = g (Rearth)2 = (32.2ft/s2) (20,908,800ft)2 Plug all of this into the W-E Theorem above and solve for vB.
|
|
|
Problem Set #9 – Problem 2 A spacecraft has an elliptical orbit. At a point A in its orbit it has a velocity of 26.9 km/s. You need to find how much to slow down the craft at point A in order to put it into an elliptical orbit that includes point B.
Point A is the point on the elliptical orbit that is the maximum distance from the planet Jupiter, and Point B is the point on the elliptical orbit that is the minimum distance from the planet.
Tangent Note: The spacecraft is moving in an elliptical orbit because it is acted upon by a single central force that is gravitational. This is the case for the stars in our solar system. They all have elliptical orbits because they are being acted upon by the single central force of the sun's gravitational field. That is, more or less. The planets are also affected by the gravitational fields of each other, however, these other forces are largely negligible compared to the gravitational force of the sun.
The fact that the single force acting is a gravitational force means you can apply the Conservation of (Mechanical) Energy.
The fact that the single force acting is a central force means you can apply the Principle of Conservation of Angular Momentum.
Applying the Conservation of (Mechanical) Energy: TA + VA = TB + VB where TA = ½ mvA2 VA = (- G m 319Mearth) / rA TB = ½ mvB2 VB = (- G m 319Mearth) / rB Note: Rearth = (6370 km) = 6370,000 m G Mearth = g (Rearth)2 = (9.81m/s2) (6,370,000 m)2
Applying the Principle of Conservation of Angular Momentum: rA mvA sinqA = rB mvB sinqB Since qA and qB are both 90o and the masses will cancel out, this equation becomes: vB = (rA/rB) vA
Substituting this expression into the Conservation of (Mechanical) Energy will allow you to solve for vA. vA= 12.7 x 103 m/s = 12.7 km/s
The speed of the spacecraft needs to be decreased by (26.9 – 12.7) km/s = 14.2 km/s
|
|
|
Problem Set #9 – Problem 3 Again, you have a body, in this case a satellite, that is acted upon by a single central force that is gravitational. Hence it will have an elliptical orbit.
You are asked to find the maximum and minimum distances of the satellite from the center of the earth in its elliptical orbit.
The fact that the single force acting is a gravitational force means you can apply the Conservation of (Mechanical) Energy.
The fact that the single force acting is a central force means you can apply the Principle of Conservation of Angular Momentum.
Applying the Conservation of (Mechanical) Energy: T0 + V0 = T1 + V1 T0 + V0 = T2 + V2 where 0 refers to the initial position, 1 is the position where the distance is maximum and 2 is the position where the distance is minimum. hence T0 = ½ m v02 T1 = ½ m v12 T2 = ½ m v22 V0 = - GMm /r0 V1 = - GMm /r1 V2 = - GMm /r2 Substituting these into the two equations for Conservation of Energy gives: ½ m v02 + - GMm /r0 = ½ m v12 + - GMm /r1 ½ m v02 + - GMm /r0 = ½ m v22 + - GMm /r2
Now Apply the Principle of Conservation of Angular Momentum: r0mv0sin (90o – a) = r1mv1sin(90o) r0mv0sin (90o – a) = r2mv2sin(90o) A bit of simplification makes these two equations look like: v1 = r0 v0 cosa / r1 v2 = r0 v0 cosa / r2
Substitute these expressions for v1 and v2 into the two equations obtained using the Conservation of (Mechanical) Energy and solve for r1 and r2.
You now have expression for r1 and r2 in terms of r0, v0 and a. However, you need to eliminate v0. Can you find v0 in terms of r0 and a? The answer, of course, is yes.
Find v0 by using the idea that the intention was to put the satellite into a circular orbit around the earth. (A circular orbit is a special case of an elliptical orbit.) However, it was projected at an angle a, not horizontally. This malfunction caused it to go into an elliptical orbit.
If the satellite had been projected horizontally, then the normal component of the satellite's acceleration, an, would have been directed towards the earth. an = v02 /r0 The force in this direction is the gravitational force: Fn = GMm/r02 Apply SFn = an: GMm/r02 = m v02 /r0 Solving for v0 gives: v02 = GM/r0 Substituting this into the expressions you found for r1 and r2 allows you to solve for r1 and r2 in terms of r0 and a only.
|