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At what height above the earth's surface the acceleration due to gravity will be 1/9th of its value at the earth's surface? A)12800 km B)2500km C)34000km D)28000km Correct answer is option 'A'.can you explain this answer.?
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To find the height above the earth's surface at which the acceleration due to gravity will be 1/9th of its value at the earth's surface, we can use the formula:
g' = (G*M)/(R+h)^2
where g' is the acceleration due to gravity at height h above the earth's surface, G is the universal gravitational constant, M is the mass of the earth, R is the radius of the earth, and h is the height above the earth's surface.
Let's substitute the given values:
g' = (1/9)*g
G*M/(R+h)^2 = (1/9)*G*M/R^2
(R+h)^2 = 9*R^2
R+h = 3*R
h = 2*R
So, the height above the earth's surface at which the acceleration due to gravity will be 1/9th of its value at the earth's surface is 2 times the radius of the earth, which is approximately 12,800 km.
Therefore, the correct answer is option A) 12800 km.
g' = (G*M)/(R+h)^2
where g' is the acceleration due to gravity at height h above the earth's surface, G is the universal gravitational constant, M is the mass of the earth, R is the radius of the earth, and h is the height above the earth's surface.
Let's substitute the given values:
g' = (1/9)*g
G*M/(R+h)^2 = (1/9)*G*M/R^2
(R+h)^2 = 9*R^2
R+h = 3*R
h = 2*R
So, the height above the earth's surface at which the acceleration due to gravity will be 1/9th of its value at the earth's surface is 2 times the radius of the earth, which is approximately 12,800 km.
Therefore, the correct answer is option A) 12800 km.
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At what height above the earth's surface the acceleration due to gravity will be 1/9th of its value at the earth's surface? A)12800 km B)2500km C)34000km D)28000km Correct answer is option 'A'.can you explain this answer.?
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