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# assuming that the gravitational potential energy of an object at infinity

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## A particle of mass m is located outside a uniform sphere of mass M at a distance r from its centre. Find the potential energy of gravitational interaction of the particle and the sphere.

Click here👆to get an answer to your question ✍️ Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by, [NEET-2019 (Odisha)] GMm GMm R+h GMmh (3) (4) mgh (1) Ruh

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## Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by, [NEET-2019 (Odisha)] GMm GMm R+h GMmh (3) (4) mgh (1) Ruh

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Updated on : 2022-09-05

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## [Solved] Assuming that the gravitational potential energy of an objec

CONCEPT: The potential energy is written as; $$U = -GMm[\frac{1}{r_f} - \frac{1}{r_i}]$$ Here we have M is the mass of earth, m is the mass of an object,&nbs

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## Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final-initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by,

mgh GMmR+h −GMmR+h GMmhR(R+h)

Option 4 : GMmhR(R+h) Crack with

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## Detailed Solution

CONCEPT:

The potential energy is written as;

U=−GMm[1rf−1ri]

Here we have M is the mass of earth, m is the mass of an object, rf

is the final radius, ri is the initial radius and G is the gravitational constant.

CALCULATION:

At initial the object will have a radius of R

In the final, the object will have a radius of R+h

Now the potential energy is written as;

U=−GMm[1rf−1ri] ⇒U=−GMm[1R+h−1R]

⇒U=−GMm[R−R−hR(R+h)]

⇒U=GMm[hR(R+h)]

Hence, option 4) is the correct answer.

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## Assuming that tha gravitational potential energy of an object at infinity is zero, the change in potential energy (final

Gravitational potential energy of the two particle system can be written as follows : U = (Gm(1)m(2))/(r ), Hence potential energies in two cases can be written as follows : (P.E.)(A) = (GMm)/(R ) (P.E.)(B) = (GMm)/(R+h) :. Delta U = (P.E.)(B) - (P.E.)(A) = (GMm)/(R+h) + (GMm)/(R ) = (GMmh)/(R(R+h)) hence option (c ) is correct.

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Assuming that tha gravitationa...

Assuming that tha gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by

Updated On: 27-06-2022

( 00 : 22 ) ADVERTISEMENT Text Solution Open Answer in App A GMm R+H GMmR+H B − GMm R+H -GMmR+H C GMmh R(R+H) GMmhR(R+H) D mgh Answer

Solution

Gravitational potential energy of the two particle system can be written as follows :

U= G m 1 m 2 r U=Gm1m2r

, Hence potential energies in two cases can be written as follows :

(P.E.) A = GMm R (P.E.)A=GMmR (P.E.) B = GMm R+h (P.E.)B=GMmR+h ∴ΔU= (P.E.) B − (P.E.) A ∴ΔU=(P.E.)B-(P.E.)A = GMm R+h + GMm R = GMmh R(R+h)

=GMmR+h+GMmR=GMmhR(R+h)

hence option (c ) is correct.

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