Question

A satellite, moving in an elliptical orbit, is 360 km above Earth’s surface at its farthest point and 180 km above at its closest point. Calculate (a) the semimajor axis and (b) the eccentricity of the orbit. c.) period of the orbit, d.) the mechanical energy of the orbiting satellite e.) the velocity of the satellite at the closest point

Answer 1

given
h1 = 360 km = 0.36*10^6 m
h2 = 180 km = 0.18*10^6 m
we know,
Re = 6.37*10^6 m
Me = 5.97*10^24 kg

a) semimajor axis, a = Re + (h1 + h2)/2

= 6.37*10^6 + (0.36 + 0.18)*10^6/2

= 6.64*10^6 m

b) Re + h1 = a*(1 + e)

6.37*10^6 + 0.38*10^6 = 6.64*10^6*(1 + e)

==> e = 0.01355

c) T = 2*pi*a^(3/2)/sqrt(G*Me)

= 2*pi*(6.64*10^6)^1.5/sqrt(6.67*10^-11*5.97*10^24)

= 5387 s (or) 1.5 hour

d) Mechanical energy of the orbiting satellite, ME = -(1/2)*G*Me*m/a

mass of the satellite should be given

e) velocity of the satellite at the closest point,

v = sqrt(G*Me/(Re + h2))

= sqrt(6.67*10^-11*5.97*10^24/(6.37*10^6 + 0.18*10^6))

= 7797 m/s (or) 7.797 km/s