Answer:
a) 2148 km = 2150 km
b) 840 km
Step-by-step explanation:
The force keeping the satellite in circular motion is the force given by Netwon's gravitational law
Centripetal force = (mv²/r)
Force due to Newton's law of gravitation = (GMm/r²)
where m = mass of satellite
M = mass of the earth
G = Gravitational constant
v = velocity of the satellite
r = radius of circular orbit
(mv²/r) = (GMm/r²)
v² = (GM/r)
Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)
v² = (k/r)
when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)
v₀² = (k/12,378)
K = 12378v₀²
When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁
1.21v₀² = (k/r₁)
r₁ = (k/1.21v₀²)
Recall, k = 12378v₀²
r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km
10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface
New altitude of orbit = 3851.75 km
Decrease in altitude = 6000 - 3851.75 = 2148 km
b) The period of orbit is related to the radius of orbit through Kepler's Law
T² ∝ R³
T² = kR³
When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)
T₀² = kR₀³
T₀² = k(12378)³
k = (T₀²) ÷ (12378)³
When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁
T₁² = kR₁³
(0.90T₀)² = kR₁³
0.81T₀² = kR₁³
R₁³ = (0.81T₀²) ÷ k
Recall, k = (T₀²)/(12378)³
R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]
R₁³ = 1,536,160,005,663.1
R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km
New Altitude = R₁ - (Radius of the Earth)
= 11,538.4 - 6378 = 5160.4 km
Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km
Hope this Helps!!!