Question

1) The small spherical planet called "Glob" has a mass of 7.72×10¹⁸ kg and a radius of 6.17×10⁴ m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 2.04×10³ 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 Nm²/kg².) (In m/s)

2) A 41.0 kg satellite is in a circular orbit with a radius of 1.60×10⁵ 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