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The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is
- a)R/√2
- b)R/2
- c)√2 R
- d)2 R
Correct answer is option 'D'. Can you explain this answer?
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The height at which the acceleration due togravity becomes g/9(where g...
To find the height at which the acceleration due to gravity becomes g/9, we need to equate the gravitational force at that height to the centripetal force required for circular motion at that height.
At a height h above the surface of the Earth, the gravitational force can be calculated using the formula:
F = (GMm) / (R+h)^2
Where:
G is the gravitational constant
M is the mass of the Earth
m is the mass of the object experiencing gravity
R is the radius of the Earth
h is the height above the surface of the Earth
The centripetal force required for circular motion at that height is given by:
F = (mv^2) / (R+h)
Where:
m is the mass of the object experiencing circular motion
v is the velocity of the object experiencing circular motion
Since we are looking for the height at which the acceleration due to gravity becomes g/9, we can set these two forces equal to each other:
(GMm) / (R+h)^2 = (mv^2) / (R+h)
Canceling out the mass (m) and velocity (v) terms, we get:
GM / (R+h)^2 = v^2 / (R+h)
Since the acceleration due to gravity at that height is g/9, we can substitute g/9 for v^2:
GM / (R+h)^2 = (g/9) / (R+h)
Cross multiplying and simplifying, we get:
9GM = g(R+h)
Dividing both sides by g, we get:
9M = R+h
Rearranging the equation, we get:
h = 9M - R
Therefore, the height at which the acceleration due to gravity becomes g/9 is 9M - R.
At a height h above the surface of the Earth, the gravitational force can be calculated using the formula:
F = (GMm) / (R+h)^2
Where:
G is the gravitational constant
M is the mass of the Earth
m is the mass of the object experiencing gravity
R is the radius of the Earth
h is the height above the surface of the Earth
The centripetal force required for circular motion at that height is given by:
F = (mv^2) / (R+h)
Where:
m is the mass of the object experiencing circular motion
v is the velocity of the object experiencing circular motion
Since we are looking for the height at which the acceleration due to gravity becomes g/9, we can set these two forces equal to each other:
(GMm) / (R+h)^2 = (mv^2) / (R+h)
Canceling out the mass (m) and velocity (v) terms, we get:
GM / (R+h)^2 = v^2 / (R+h)
Since the acceleration due to gravity at that height is g/9, we can substitute g/9 for v^2:
GM / (R+h)^2 = (g/9) / (R+h)
Cross multiplying and simplifying, we get:
9GM = g(R+h)
Dividing both sides by g, we get:
9M = R+h
Rearranging the equation, we get:
h = 9M - R
Therefore, the height at which the acceleration due to gravity becomes g/9 is 9M - R.
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The height at which the acceleration due togravity becomes g/9(where g = the accelerationdue to gravity on the surface of the earth) in terms of R, the radius of the earth, isa)R/√2b)R/2c)√2 Rd)2 RCorrect answer is option 'D'. Can you explain this answer?
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