Question

Problem 3: You and your friend are both starship captains. Your friend is orbiting a planet in a circular orbit at an altitude of 2 x 10 m above the surface. Your ship is also in a circular orbit but is at an altitude of 7 x 106m above the surface. (Mplanet = 2 x 1023 kg and Rplanet = 4 x 100m) a) [4 points) What are the orbital speeds of the two ships in their circular orbits? b) [6 points) By what amount would you Diagram for parts B&C need to alter your velocity to drop your orbit down so that it touches your Your friend's orbit as shown in the diagram to the right? Would this be an increase or decrease in velocity? Friend ci

c.) How fast would you be moving at point "C" in the provided diagram?

d.) At point C, you determine that the center of the planet's moon (Mass of Moon = 6 x 10^22) is 4 x 10^7 meters away from you. What is the escape velocity for the planet/ Moon system at this point?

Answer 1

a)The equation of orbital velocity

•   $v=\sqrt{\frac{G\times M}{r}}$
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  6.674x10ll m3 kg-1 sec1 orbital velocity of satellite Vy GIMP Mp= 24 20 23 kg distance from planet sugace + R planet orbital (friend's) - 6 • 674x10 11 x ax 1023 68106 Velocity 1.4915 km's = 1491-53 ms Vo (yours) 6.67 4 x 10-11 x 2 x 1023 106 1.101 Kms 1101.569 ms -
  
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  (6) The amount at which the velocity should be changed to touch friends og bit. AV = Vfriends - Vyours 0.39 05 km/s = 1.4915 - 1.101 - You need to increace the velocity by 0.3905 km/s to touch friends orbit. (0) In the provided diagram c is a point on the friends orbit. So you need travel same speed as of friends. At point a your orbital speed Vo = 104915 km/s
  
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  DM SY-Distance blu planet and moon. Þ CO for the satellite to escape from the moon-planet System. The kinetic energy of satellite should be equal to Gravitational Field potential energy. (distance blw planet and moon). Dm 6 * to Mm= 6x 1020 kg. Dm = 4x107m At poont c, the for escape Velocity of Starship can be found out from energy equation: Y2 mp v2 = G Mp or GMMM DM op 12 2 GMP Exp + GMm DM v2= 2 GMP + 2 G Mm Тр DM
  
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  GMP + GMm Dm Вр + 6°674410 x 6x1022 6. 67410 x 2x 1023 11 68106 4x107 2.224 X106 + 0.101 X10 06) = 2.15 63 km/s V 4065 at escape Velocity of planet 2015 63 kms C =