Question

(TWO PARTS) The small spherical planet called "Glob" has a mass of 7.46×1018 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 1.56×103 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)The small spherical planet called "Glob" has a mass of 7.46×1018 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 1.56×103 m, above the surface of the planet, before it falls back down.

a) 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)

b) A 28.0 kg satellite is in a circular orbit with a radius of
1.55×10^{5} m around the planet Glob. Calculate the speed
of the satellite.

(in m/s)

Answer #1

a)

Use the formula for gravitational potential energy (PE =
-GMm/r); and use the principle of conservation of energy.

Let:

M = mass of Glob

m = mass of rock (unknown)

r = radius of Glob

h = maximum height of rock

v = rock's initial speed

At the moment the rock is thrown, it has this much energy:

E_initial = KE_initial + PE_initial

E_initial = ½mv² + (-GMm/r)

When it reaches its maximum height h, it has this much
energy:

E_final = KE_final + PE_final

E_final = 0 + -GMm/(r+h)

By conservation of energy:

E_initial = E_final

½mv² - GMm/r = -GMm/(r+h)

Divide by the rock's mass (m), and it disappears:

½v² - GM/r = -GM/(r+h)

GMm ( 1/r – 1/(r+h) ) =
6.67*10^{–11}*7.46*10^{18} m ( 1/64100 – 1/65660 )
= m*184.42

m*184.42 = 1/2 m v^2

v = 19.20 m/s

(b)

GMm/R^2=m x v^2/R

v^2=GM/R

v^2=(6.67x10^{-11} x 7.46×10^{18}) /
1.55x10^{5}m

v=56.65 m/s

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