Question

1) The small spherical planet called "Glob" has a mass of
7.72×10^{18} kg and a radius of 6.17×10^{4} m. An
astronaut on the surface of Glob throws a rock straight up. The
rock reaches a maximum height of 2.04×10^{3} m, above the
surface of the planet, before it falls back down. What was the
initial speed of the rock as it left the astronaut's hand? (Glob
has no atmosphere, so no energy is lost to air friction. G =
6.67×10^{-11} Nm^{2}/kg^{2}.) (In
m/s)

2) A 41.0 kg satellite is in a circular orbit with a radius of
1.60×10^{5} m around the planet Glob. Calculate the speed
of the satellite.

(in m/s)

Answer #1

Given that,

Mass of Glob =M = 7.72 x 10^18 Kg ; Radius of the Glob = r = 6.17 x 10^4 meters

Let m be the mass of rock

Let v be the initial velocity to be calculated.

Max. height reached = H = 2.04 x 10^3 meters

Energy is conserved everywhere, So in this case at max height the KE of rock equals to PE. So we can write

1/2 m v2 + G m M/R = G m M H / (R + h)

v = sqrt [2 G M h/R(R + h)]

v = sqrt [2 x 6.67 x 10^-11 x 7.72 x 10^18 x 2.04 x 10^3/[6.17 x 10^4(6.17 x 10^4 + 2.04 x 10^3)]] = 23.11 m/s

**Hence,intial speed of the rock = v = 23.11
m/s**

b)v = sqrt (GM/r)

v = sqrt (6.67 x 10^-11 x 7.72 x 10^18/(1.6 x 10^5)) = 56.73 m/s

**Hence, v = 53.76 m/s**

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