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The mean radius of earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite ? [1992]
- a)( R2g/ω2)1/3
- b)( Rg/ω2)1/3
- c)( R2ω2/g)1/3
- d)( R 2g /ω)1/3
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The mean radius of earth is R, its angular speed on its own axis is &o...
Hence, 
Most Upvoted Answer
The mean radius of earth is R, its angular speed on its own axis is &o...
GMm/r² ; m omega²r
r³ = GM/omega² = gR²/omega²
so, r = (gR²/omega²)⅓
r³ = GM/omega² = gR²/omega²
so, r = (gR²/omega²)⅓
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The mean radius of earth is R, its angular speed on its own axis is &o...
Concept:
The radius of the orbit of a geostationary satellite can be calculated using the formula for centripetal force in a circular motion.
Given:
Mean radius of Earth = R
Angular speed on its own axis = ω
Acceleration due to gravity at Earth's surface = g
Formula:
For a geostationary satellite, the centripetal force required for it to stay in orbit is provided by the gravitational force.
So, equating the centripetal force to the gravitational force:
m * v^2 / r = G * M * m / r^2
Where,
m = mass of the satellite
v = velocity of the satellite
r = radius of the orbit
G = universal gravitational constant
M = mass of the Earth
Solution:
- We know that v = r * ω (where ω is the angular velocity)
- Equating the centripetal force to the gravitational force, we get:
m * (r * ω)^2 / r = G * M * m / r^2
- Simplifying, we get:
r^3 = G * M / ω^2
- Substituting the values of G, M, and ω, we get:
r^3 = g * R^2 / ω^2
- Taking the cube root on both sides, we get:
r = (R^2 * g / ω^2)^(1/3)
Therefore, the radius of the orbit of a geostationary satellite is given by option 'A' i.e., (R^2 * g / ω^2)^(1/3).
The radius of the orbit of a geostationary satellite can be calculated using the formula for centripetal force in a circular motion.
Given:
Mean radius of Earth = R
Angular speed on its own axis = ω
Acceleration due to gravity at Earth's surface = g
Formula:
For a geostationary satellite, the centripetal force required for it to stay in orbit is provided by the gravitational force.
So, equating the centripetal force to the gravitational force:
m * v^2 / r = G * M * m / r^2
Where,
m = mass of the satellite
v = velocity of the satellite
r = radius of the orbit
G = universal gravitational constant
M = mass of the Earth
Solution:
- We know that v = r * ω (where ω is the angular velocity)
- Equating the centripetal force to the gravitational force, we get:
m * (r * ω)^2 / r = G * M * m / r^2
- Simplifying, we get:
r^3 = G * M / ω^2
- Substituting the values of G, M, and ω, we get:
r^3 = g * R^2 / ω^2
- Taking the cube root on both sides, we get:
r = (R^2 * g / ω^2)^(1/3)
Therefore, the radius of the orbit of a geostationary satellite is given by option 'A' i.e., (R^2 * g / ω^2)^(1/3).
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