Question Details

A body placed at a distance R₀ form the center of earth, starts moving from rest. The velocity of the body on reaching at the earth’s surface will be ( Rₑ = radius of earth and Mₑ = mass of earth)

Options

A

GMₑ(1/Rₑ - 1/R₀)

B

2GMₑ(1/Rₑ - 1/R₀)

C

GMₑ√(1/Rₑ - 1/R₀)

D

√(2GMₑ[1/Rₑ - 1/R₀])

Correct Answer :

√(2GMₑ[1/Rₑ - 1/R₀])

Solution :

The correct option is:

2GMe1Re-1R0

Step-by-Step Explanation:

To find the velocity of the body when it reaches the Earth's surface, we can use the law of conservation of mechanical energy. Since there are no non-conservative forces acting on the body (such as air resistance), the total mechanical energy of the body remains constant throughout its motion.

1. Initial State (at distance R0 from the center of the Earth):
The body starts from rest, which means its initial velocity is zero. Thus, its initial kinetic energy is:

Ki=0

The gravitational potential energy of a body of mass m at a distance R0 from the center of the Earth (mass Me) is:

Ui=-GMemR0

Therefore, the initial total mechanical energy (Ei) is:

Ei=Ki+Ui=-GMemR0

2. Final State (on reaching the Earth's surface at distance Re from the center):
Let the velocity of the body upon reaching the surface be v. The final kinetic energy is:

Kf=12mv2

The final gravitational potential energy at the Earth's surface (distance Re) is:

Uf=-GMemRe

Therefore, the final total mechanical energy (Ef) is:

Ef=Kf+Uf=12mv2-GMemRe

3. Applying Conservation of Energy:
Equating the initial total mechanical energy to the final total mechanical energy:

Ei=Ef

-GMemR0=12mv2-GMemRe

We can cancel out the mass of the body (m) from both sides of the equation:

-GMeR0=12v2-GMeRe

Now, rearrange the terms to solve for v2:

12v2=GMeRe-GMeR0

Factor out the common term GMe on the right-hand side:

12v2=GMe1Re-1R0

Multiply both sides by 2:

v2=2GMe1Re-1R0

Taking the square root of both sides gives the final velocity of the body on reaching the Earth's surface:

v=2GMe1Re-1R0

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