Question

1. A certain satellite travels in an approximately circular orbit of radius 7.5 × 10⁶ m with a period of 6 h 27 min. Calculate the mass of its planet from this information

2. (a) At what height above Earth's surface is the energy required to lift a satellite to that height equal to the kinetic energy required for the satellite to be in orbit at that height? (b) For greater heights, which is greater, the energy for lifting or the kinetic energy for orbiting?

Please help:)

Answer 1

Here ,

let the mass of planet is M

1)

radius of orbit , r = 7.5 *10^6 m

time period of planet , T = 6 h 27 min

as the period of planet is given as

T^2/r^3 = 4*pi^2/(G * M)

(6 * 3600 + 27 * 60)^2/(7.5 *10^6)^3 = 4*pi^2/(6.673 *10^-11 * M)

solving for mass M

M = 4.62 *10^23 Kg

the mass of planet is 4.62 *10^23 Kg

2)

let the height is h

R is the radius of earth

M is the mass of earth

m is the mass of satellite

as the kinetic energy of a satellite is

for kinetic energy = energy needed to raise the satellite

G * M * m /(2 * r) = G *M * m (1/R - 1/(R + r))

1/(2 * r) = 1/R - 1/(R + r)

r = R

when the height of satelite is equal to radius of earth , thn Earth's surface is the energy required to lift a satellite to that height equal to the kinetic energy required for the satellite to be in orbit at that height

b) at higher heights ,

as the change in potential energy is more

the energy for lifting the satellite is more