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The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is V. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is
- a)3/2 V
- b)√3/2 V
- c)√2/3 V
- d)2/3 V
Correct answer is option 'C'. Can you explain this answer?
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√2 Vc)√3/2 Vd)√2/3 Ve) V/2
We can use the formula for orbital velocity:
V = √(GM/R)
where G is the gravitational constant, M is the mass of the earth, and R is the distance from the center of the earth to the satellite. For a circular orbit just above the earth's surface, R is equal to the radius of the earth (denoted by Re). Therefore,
V = √(GM/Re)
For a satellite orbiting at an altitude of half of the earth's radius (denoted by 0.5Re), the distance from the center of the earth is R = Re + 0.5Re = 1.5Re. So, the orbital velocity for this satellite is:
V' = √(GM/(1.5Re))
Dividing V' by V, we get:
V'/V = √(GM/(1.5Re)) / √(GM/Re)
Simplifying, we get:
V'/V = √(Re/(1.5Re))
V'/V = √(2/3)
Therefore, the answer is (d) √2/3 V.
We can use the formula for orbital velocity:
V = √(GM/R)
where G is the gravitational constant, M is the mass of the earth, and R is the distance from the center of the earth to the satellite. For a circular orbit just above the earth's surface, R is equal to the radius of the earth (denoted by Re). Therefore,
V = √(GM/Re)
For a satellite orbiting at an altitude of half of the earth's radius (denoted by 0.5Re), the distance from the center of the earth is R = Re + 0.5Re = 1.5Re. So, the orbital velocity for this satellite is:
V' = √(GM/(1.5Re))
Dividing V' by V, we get:
V'/V = √(GM/(1.5Re)) / √(GM/Re)
Simplifying, we get:
V'/V = √(Re/(1.5Re))
V'/V = √(2/3)
Therefore, the answer is (d) √2/3 V.
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The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is V. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity isa)3/2 Vb)√3/2 Vc)√2/3 Vd)2/3 VCorrect answer is option 'C'. Can you explain this answer?
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