Ten days after being launched to Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87*10^6 km from Earth and moving at 1.20*10^4 km/ h with respect to the Earth.
A. Kinetic energy = 0.5 * m * v^2
= 0.5 * 629 kg * ((1.2 * 10^7 m/h)/(3600 s/h))^2 kg-m/s
For the second question (b),
You must use the potential energy formula U = -G*M*m/R
G = gravitational constant = 6.67 *10^-11
M = mass of the earth = 5.98*10^24 kg
m = mass of satellite (object) = 629 kg
R = the distance between the earth and the object = 2.87*10^6 km (remember to change this to meters by multiplying by 1000)
plug them all in and your answer should be -8.7417*10^7 Joules
The kinetic energy of a body is given by
J=1/2*mass*velocity^2
then
1/2 *629Kg *(1.20*10^4km/h)^2
This provides the spacecraft’s kinetic energy relative to Earth.
The second is more difficult, and I will have to research some things…
a) Convert the speed to meters/second and calculate E = mv^2/2. The mass of the satellite is m.
b) Evaluate PE = m(rs/r – 1) where r is the radius of the Earth and rs is the distance from the satellite to the Earth.
potential energy is MAD (mass x acceleration x distance)