Question

a)The small spherical planet called "Glob" has a mass of 8.42×1018 kg and a radius of 6.41×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.96×103 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.) b)A 35.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.

Answer 1

a)

M = mass of planet = 8.42 x 10¹⁸ kg

m = mass of rock

v = initial speed

r = radius of planet = 6.41 x 10⁴ m

h = height reached

using conservation of energy

KE + PE at the surface = PE at the top

(0.5) m v² - GMm/r = - GMm/(r+ h)

(0.5) v² - GM/r = - GM/(r+ h)

(0.5) v² = GM/r - GM/(r+ h)

(0.5) v² = (6.67 x 10^-11) (8.42 x 10¹⁸)(1/(6.41 x 10⁴) - 1/((8.42 x 10¹⁸) + (1.96 x 10³))

v = 132.4 m/s

b)

r = radius of orbit = 1.45 x 10⁵ m

orbital speed is given as

v = sqrt(GM/r)

v = sqrt((6.67 x 10^-11) (8.42 x 10¹⁸ )/(1.45 x 10⁵ )) = 62.2 m/s