Question

The small spherical planet called "Glob" has a mass of
7.56×10^{18} kg and a radius of 6.44×10^{4} m. An
astronaut on the surface of Glob throws a rock straight up. The
rock reaches a maximum height of 1.36×10^{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}.)

A 39.0 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.

Answer #1

First we need to find the change in Potential energy. So we will
get the potential energy there. This potential energy is equal to
the Kinetic energy. By equating that and we can solve for
v,velocity.

GMm ( 1/r – 1/R ) = 6.67 . 10^{–11} *7.56 . 10^{18}
m ( 1/(64400)- 1/(64400+ 1360 ) = m 161.93J

161.93m = 1/2 m v^{2}

v = 17.99 m/s

b)So, GMm/R^{2}=mv^{2}/R

Sice it is a circular orbit a
centripetal Force will be there, mv^{2}/R is the
centripetal force

Simplifying, v^{2}=GM/R

v^{2}=(6.67x10^{-11} x 7.56x10^{18}) /
1.60x10^{5}

The mass of the satelite is cancelled,

Solving,v=56.13 ms^{-1}

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