# If 'g' is the acceleration due to gravity on earth then increase in P.E of a body of mass 'm' up to a distance equal to the radius of earth from the earth surface will be ?

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Now, when the body is on the surface, the PE of the system

Once more, the system's PE at height h from the earth's surface is provided by

When h equals R, then

#E_R=-(GmM)/(R+R)=-(GmM)/(2R) ....(3)#

Therefore, the increase in PE caused by the body of mass m shifting from the surface to a height equal to the earth's radius (R) is as follows:

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The increase in potential energy (P.E) of a body of mass 'm' when raised to a distance equal to the radius of the Earth from the Earth's surface is equal to mgR, where 'g' is the acceleration due to gravity on Earth and 'R' is the radius of the Earth.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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