Question

The small spherical planet called "Glob" has a mass of 6.50×10¹⁸ kg and a radius of 5.60×10⁴ m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.64×10³ 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)

-A 33.0 kg satellite is in a circular orbit with a radius of 1.40×10⁵ m around the planet Glob. Calculate the speed of the satellite.
(in m/s)

Answer 1

First you want to know the acceleration due to gravity of the planet Glob. At the surface the gravitational acceleration (g) can be determined by equating the weight of an object with the gravitational force law

m*g = G*M*m/R^2

where R is the radius of Glob, M is Glob's mass, G is the universal gravitational constant, and m is the mass of the object. Solving for g yields

g = G*M/R^2

Now, throwing an object into the air, and neglecting air resistance, we can consider conservation of energy. The kinetic energy of the object when it is released by the astronaut is completely converted into potential energy when the object reaches the top of its travel.

0.5*m*V^2 = m*g*H

where V is the initial speed of the object and H is the height to which it traveled, and g is the gravitational acceleration solved for above. Solving for V we find

V = sqrt(2*g*H).
V=sqrt(2*9.81*1.64*10^3)

V=179.38 m/s

2.The satellite traveling in a circular orbit of radius D above the planet has its centrifugal force balanced by the gravitational force of the planet.

Fc = m*v^2/D = G*M*m/D^2 = Fg

Solving for the speed of the satellite (v) needed in order to maintain orbit at radius D yields

v = sqrt(G*M/D).

v=sqrt(6.67*10^-11.*1.4*10^5*33)

V=0.012529 m/s

in secound i am not sure