Question

A spacecraft is in a circular Earth orbit at an altitude of 5400 km . By how much will its altitude decrease if it moves to a new circular orbit where (a) its orbital speed is 10% higher or (b) its orbital period is 10% shorter

Answer 1

Answer:

Part (a) the decrease in altitude is 2048 km

Part (b) the decrease in altitude is 818.2 km

Explanation:

Part (a) when its orbital speed is 10% higher

$V = \sqrt{\frac{gR_e^2}{r} }$

where;

V is the orbital speed m/s

g is acceleration due to gravity = 9.8 m/s

Re is the radius of Earth = 6400 km

r is the resultant distance of the spacecraft from center of Earth

r = altitude of the spacecraft + Re = 5400km + 6400km = 11800 km

$V_1 = \sqrt{\frac{9.8*(6400000)^2}{(11800000)}} = 5832.46m/s$

V₂ = 1.1V₁ = 6415.71 m/s

$r_2 = \frac{gR_e^2}{V_2^2} = \frac{9.8*(6400000)^2}{(6415.71)^2} = 9752.1 km$

Altitude = 9752.1 km - 6400 km = 3352.1 km

Decrease in altitude = 5400 km -3352.1 km = 2048 km

Part (b) when its orbital period is 10% shorter

$T_1 = 2\pi r\sqrt{\frac{r}{gR_e^2}} = 2\pi (11800000)\sqrt{\frac{11800000}{9.8*(6400000)^2}}\\\\T_1 =12713.53 s$

But T₂ is 10% shorter = 0.9 *T₁ = 0.9*12713.53 = 11442.18s

$r = (\frac{T^2*g*R_e^2}{4\pi^2 })^{\frac{1}{3}} \\\\r = (\frac{(11442.18)^2*9.8*(6400000)^2}{4\pi^2 })^{\frac{1}{3}}\\\\r = (1.3309*10^{21})^{\frac{1}{3}} = 10981.8 km$

Altitude = 10981.8 km - 6400 km = 4581.8 km

Decrease in altitude = 5400 km - 4581.8 km = 818.2 km