Question

The small spherical planet called "Glob" has a mass of 7.56×10¹⁸ kg and a radius of 6.44×10⁴ m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.36×10³ 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.)

A 39.0 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.

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 . 10¹⁸ m ( 1/(64400)- 1/(64400+ 1360 ) = m 161.93J

161.93m = 1/2 m v²

v = 17.99 m/s

b)So, GMm/R²=mv²/R

Sice it is a circular orbit a centripetal Force will be there, mv²/R is the centripetal force

Simplifying, v²=GM/R

v²=(6.67x10^-11 x 7.56x10¹⁸) / 1.60x10⁵

The mass of the satelite is cancelled,

Solving,v=56.13 ms^-1