Question

A 500-kg satellite is in a circular orbit at an altitude of 500 km above the Earth’s surface. Because of air friction, the satellite eventually falls to the Earth’s surface, where it hits the ground with a speed of 2.00 km/s. How much energy was transformed into internal energy by means of air friction?

Answer 1

$\text{Let}\newline M=6*10^{24}kg\text{ Earth mass}$

$R = 6.4*10^7 m \text{ radius of the Earth}$

$G= 6.7*10^{-11}\text{ gravitational constant}$

$h = 0.5*10^7 m \text{ altitude of satellite }$

$m= 5*10^2\ kg\text{ satellite mass}$

$V\text{ satellite speed in orbit }$

$V_1=2*10^3m/s\text{ satellite landing speed}$

$\text{in orbit:}$

$F=ma$

$F= G\frac{mM}{(R+h)^2}$

$ma= G\frac{mM}{(R+h)^2}$

$a= G\frac{M}{(R+h)^2}$

$a=\frac{V^2}{(R+h)}$

$V^2= G\frac{M}{(R+h)}$

$E_k=\frac{mV^2}{2}=G\frac{mM}{2*(R+h)}$

$E_p=-G\frac{mM}{R+h}$

$E=E_k+E_p=-G\frac{mM}{2(R+h)}=-0.29*10^{10}$

$\text{upon landing:}$

$E_l=E_{p1}+E_{k1}$

$E_l=-G\frac{mM}{R}+\frac{mV_1^2}{2}=-0.30*10^{10}$

$Q= E-E_l=0.1*10^{10}$

Answer:$0.1*10^{10}J -\text{transformed into internal energy by means of air friction}$