Question

A satellite of mass 2500 kg is orbiting the Earth in an elliptical orbit. At the farthest point from the Earth, its altitude is 3600 km, while at the nearest point, it is 1100 km. Calculate the energy and angular momentum of the satellite and its . at the aphelion and perihelion.

Answer 1

GRAVITATIONAL POTENTIAL ENERY:
U=-GMm/r
where  G= gravitational constant=6.67x10⁻¹¹ Nm/kg²
M=earth mass
m=mass of planet.
Aphelion is a point in orbit  where satellite is  farthest from earth
r1=RE +3600km=9.96 x 10⁶ m
Perihelion is a point in orbit where satellite is nearest from earth
r2=RE=1100km=7.46 x 10⁶m
According to conservation of energy,
K₁ + V1=K₂ +V₂
1/2mV₁² - GmM/r₁ =1/2 .m.V₂² - G M.m/r₂
According to conservation of angular  Momentum:
mv₁r₁=mv₂r₂
⇒v₂=(r1/r2).v1
v₁²/2 -GM/r₁=1/2[(r1/r2)v₁]² - GM/r²
v1²=GM(1/r1-1r2)/[1-(r1/r2)²]
Speed at Aphelion:

v1=√6.67x10⁻¹¹x5.98x10²⁴x[(1/6.63x10₆ ) -(1/6.37x10⁶+ 1.1x10⁶]
                                   --------------------------------------------------
                                   (1-(6.37x10⁶ +3.6 x 10⁶)/(6.37x10⁶+1.1x10⁶]
=4140.5m/s
speed at perihelion:
V2= 4140.5 x [6.67x10⁶+3.6 x 10⁶]/(6.37 x 10⁶ +1.1 x 10⁶]
=5526.2m/s
Thus,
energy at aphelion:
Eₐ=1/2 x [2500(4140.5)² - [6.673 x 10 ⁻¹¹ x 5.98 x 10²⁴ x2500]/[6.673 x 10⁶+3.6 x 10⁶]
=-786 x 10¹⁰J
energy At perihelion : Eₐ=Ep=-7.86 x 10¹⁰ J
Angular momentum at  aphelion:
Lₐ=2500(4140.5)(6.37x10⁶ + 3.6 x 10 ⁶]
=1.03 x 10¹⁴ kgm²/s
At Perihelion:
Lp=La=1.03 x 10¹⁴ kgm²/s