
The value of g at a particular point is 9.8 m/$sec^2$ suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will become
A. 9.8 m/$sec^2$
B. 4.9 m/$sec^2$
C. 19.6 m/$sec^2$
D. 3.1 m/$sec^2$
Answer
584.4k+ views
Hint: If we equate the magnitude of force due to gravity on the body kept at that point with mg, with m being the mass of the body, we will get the formula for g at that point due to earth's gravity.
Formula used:
If r is the separation of the point from the center of the earth then,
$g = \dfrac{GM}{r^2}$
Complete answer:
We know that gravitational force acting between two bodies is given as:
$F = \dfrac{GMm}{r^2}$
where r is the separation between the two bodies. If one of the bodies is extremely small with a mass m and another one is earth with a mass M, we can say that the weight of the body is nothing but the mass of the body m times the gravitational acceleration due to earth on that body.
Therefore, equating these two forces on the body (as they are same), we get:
$mg = \dfrac{GMm}{r^2} $ ;
Which gives us the formula for g as:
$g = \dfrac{GM}{r^2} $
Now, we are asked to find what happens when the earth shrinks, and we are given that mass M of the earth remains the same and the distance from the center of the earth to that point (body) remains the same. As from the formula, we can see that none of the parameters are getting affected by these changes; therefore new g will be the same as the old one.
So, the correct answer is “Option A”.
Note:
This question teaches an important concept that the distances of separation of two bodies are taken from their center. When the earth shirked, its radius got reduced to R' (which was R before). Initially, we had r = R + h, later we had r = R' + h'. The height of the point above the earth surface got increased and earth's radius decreased. But none of this affected the separation of the point from the center of the earth which was r. Gravity is a central force and acts along the line joining the centre of the two masses.
Formula used:
If r is the separation of the point from the center of the earth then,
$g = \dfrac{GM}{r^2}$
Complete answer:
We know that gravitational force acting between two bodies is given as:
$F = \dfrac{GMm}{r^2}$
where r is the separation between the two bodies. If one of the bodies is extremely small with a mass m and another one is earth with a mass M, we can say that the weight of the body is nothing but the mass of the body m times the gravitational acceleration due to earth on that body.
Therefore, equating these two forces on the body (as they are same), we get:
$mg = \dfrac{GMm}{r^2} $ ;
Which gives us the formula for g as:
$g = \dfrac{GM}{r^2} $
Now, we are asked to find what happens when the earth shrinks, and we are given that mass M of the earth remains the same and the distance from the center of the earth to that point (body) remains the same. As from the formula, we can see that none of the parameters are getting affected by these changes; therefore new g will be the same as the old one.
So, the correct answer is “Option A”.
Note:
This question teaches an important concept that the distances of separation of two bodies are taken from their center. When the earth shirked, its radius got reduced to R' (which was R before). Initially, we had r = R + h, later we had r = R' + h'. The height of the point above the earth surface got increased and earth's radius decreased. But none of this affected the separation of the point from the center of the earth which was r. Gravity is a central force and acts along the line joining the centre of the two masses.
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