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Problem 3: You and your friend are both starship captains. Your friend is orbiting a planet in a circular orbit at an altitud
c) [4 points) How fast would you be moving at point “C” in the provided diagram? d) [4 points) At point C, you determine that
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a) Let the speed of your's to be v1 and that of your friend's to be v2.

Now balancing the forces along radial direction;

GMplanetm1/(Rplanet + 7 x 106m)2 = m1v12/(Rplanet + 7 x 106m)

=> v1 = [(GMplanet)/(Rplanet + 7 x 106 m)]1/2

=> v1 = 1101.6 m/s

similarly;

GMplanetm2/(Rplanet + 2 x 106m)2 = m2v22/(Rplanet + 2 x 106m)

=> v2 = [(GMplanet)/(Rplanet + 2 x 106 m)]1/2

=> v2 = 1491.563 m/s

(b) Let the altered velocity of your spacecraft to be v; and vf to be the velocity when you reach your friend's orbit

Now if you have to enter your friend's orbit then you have to conserve the energy and the angular momentum to adjust the velocity; thus

m1v(Rplanet + 7 x 106m) = m1vf/(Rplanet + 2 x 106m)

=> 11v = 6vf ..........(i)

By energy conservation;

m1v2/2 - GMplanetm1/(Rplanet + 7 x 106m) = m1vf2/2 - GMplanetm1/(Rplanet + 2 x 106m)

=> v2 = 2GMplanet/(Rplanet + 7 x 106m) + vf2 - 2GMplanet/(Rplanet + 2 x 106m)

From equation (i);

=> v2 - (11v/6)2 = 2GMplanet/(Rplanet + 7 x 106m) - 2GMplanet/(Rplanet + 2 x 106m)

=> v2 = 856592.0856

=> v = 925.5226 m/s

You have to decrease your velocity.

c) From equation (i)

vf = 11v/6 = 1696.8 m/s will be the velocity of you at C.

d) The total energy of your spaceship at this particular moment has to be equal to the kinetic energy when you have escaped;

Assuming escape velocity to be ve;

m1ve2/2 - Gm1Mplanet/(Rplanet + 2 x 106m) - Gm1Mmoon/(40 x 106m) = 0

=> ve2 = 4649748.4

=> ve = 2156.33 m/s

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