A Long Fall Earth II is a planet in a distant solar system which is earth-like,

Question

A Long Fall Earth II is a planet in a distant solar system which is earth-like, but a little smaller than our earth. In the distance future, Earth II has been settled by bold travelers from our home planet. Very far from Earth II (effectively at R=), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of Earth II were to act on the spacecraft (i.e., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into Earth II. The mass of Earth II is Me=4.240×1024 kg and its radius is Re=5.700×106 m . Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space.

Part A

Find the speed se of the spacecraft when it eventually crashes into Earth II.

Part B

Now find the spacecraft's speed when its distance from the center of Earth II is 13.5 times the radius of Earth II, i.e. R=Re, where = 13.5.

Part C

Now find the spacecraft's speed when its distance from the center of Earth II is 135 times the radius of Earth II, i.e R=Re, where = 135.

Explanation / Answer

speed of a spacecraft at which it would crash into earth, assuming it were a distance of away, and no other astronomical object's gravitational pull affected it is

part A

G =6.754×1011 m3 kg1 s2

v= sqrt(( 2*Me*G)/Re) = sqrt( (4.24*1024*6.754×1011)/(5.7*106)) =sqrt(485.32) = 22.03 m/sec

part B

now v1= sqrt(( 2*Me*G)/(alpha*Re)) {let v1= speed of air craft}

therefore v1 = v/sqrt(alpha)

given alpha =13.5

v1 = 22.03/sqrt(alpha) = 5.9958 m/sec

partC
now alpha is 135

therefore
v1 {speed of air craft in this case} = v/sqrt(alpha)

= 22.03/(sqrt(135) = 1.896 m/sec

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