a satellite revolves around the earth in a circular orbit once 90 minutes. If the radius of earth is 6400 km, how high is the satellite above the earth? (g= 9.8 metre per second square)
Answers
Answered by
25
m = mass of satellite. R = Radius of Earth.
d = R + h = radius of orbit.
h = altitude of satellite above surface of Earth.
M = mass of Earth.
g = GM/R². So G M = g R²
Centripetal force = Gravitational force
m v²/d = G Mm/d² = m g R²/d²
v² = g R²/d
v = R √(g/d)
T = 2πd / v
T² = 4 π² d³ / (g R²)
d³ = g R² T² /(4π²)
= 9.8 * 6.4² * 10¹² * 90² * 60² /4π² m³
d = 6, 668.14 km
altitude h = d - R = 168.14 km
d = R + h = radius of orbit.
h = altitude of satellite above surface of Earth.
M = mass of Earth.
g = GM/R². So G M = g R²
Centripetal force = Gravitational force
m v²/d = G Mm/d² = m g R²/d²
v² = g R²/d
v = R √(g/d)
T = 2πd / v
T² = 4 π² d³ / (g R²)
d³ = g R² T² /(4π²)
= 9.8 * 6.4² * 10¹² * 90² * 60² /4π² m³
d = 6, 668.14 km
altitude h = d - R = 168.14 km
kvnmurty:
:-) :-)
271km should be the answrr
Answered by
20
h=(gR^2T^2÷4π^2)^1/3-R
=9.8×(6400)^2×(90)^2÷4×(3.14)^2)-6400
=268km
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