Question

A satellite is in elliptic orbit around the earth with aphelion of 6R and perihelion of 2 R where R = 6400 km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R?

[G = 6.67 × 10^-11 SI unit and M = 6 × 10²⁴ kg.]

Answer 1

Given

Radius or perihelion, r_p = 6R

Radius of aphelion, r_a = 2 R

Hence,

r_a = a(1+e) = 6R ...............(1)

r_p = a(1-e) = 2R ..............(2)

From (I) & (II)

eccentricity, e = \(\frac{1}{2}\)

From law of conservation of angular momentum, angular momentum oat perigee = angular momentum at apogee

\(\therefore\) mv_pr_p = mv_ar_a

\(\therefore\) \(\frac{v_a}{v_p}=\frac{1}{3}\)

Applying law of Conservation of Energy :

Energy at perigee = Energy at apogee

(Where M is mass of earth)

\(\frac{1}{2}\)\(mv^2_p\) - \(\frac{GMm}{r_a}\) = \(\frac{1}{2}mv^2_a-\frac{GMm}{r_a}\)

(where M is the mass of a earth)

\(\therefore\) \(v^2_p\big(1-\frac{1}{9}\big)\)= -2GM\(\big[\frac{1}{r_a}-\frac{1}{r_p}\big]\)= 2GM\(\big[\frac{1}{r_a}-\frac{1}{r_p}\big]\)

Putting v_a = \(\frac{v_p}{3}\)

v_p = \(\frac{2GM\big[\frac{1}{r_p}-\frac{1}{r_a}\big]^{\frac{1}{2}}}{\big[1-\big(\frac{v_a}{v_p}\big)^2\big]^{\frac{1}{2}}}\) = \(\Bigg[\frac{\frac{2GM}{R}\big[\frac{1}{2}-\frac{1}{6}\big]}{\big(1-\frac{1}{9}\big)}\Bigg]^{\frac{1}{2}}\)

= \(\Bigg(\frac{\frac{2}{3}}{\frac{8}{9}}\frac{GM}{R}\Bigg)^{\frac{1}{2}}\)

= \(\sqrt{\frac{3}{4}\frac{GM}{R}}\) = 6.85 \(\frac{km}{s}\)

v_p = 6.85 km/s

v_a = \(\frac{v_p}{3}\)

v_e = 2.28 km/s

For circular orbit of radius ,

For r = 6R, v_e = \(\sqrt{\frac{GM}{6R}}\) = 3.23 km/s.

Hence to transfer to a circular orbit at apogee, we have to boost the velocity by ∆ = (3.23 – 2.28) = 0.95 km/s. this can be done by suitably firing rockets from the satellite.