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

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

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

V₂ = 1.1V₁ = 6415.71 m/s

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

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

Altitude = 10981.8 km - 6400 km = 4581.8 km

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

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6