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Gravitation question

2024 · 6 Apr · Shift 1 · Q68
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  5. /2024 · 6 Apr · Shift 1 · Q68

Gravitation question

2024 · 6 Apr · Shift 1 · Q68

JEE MainPhysicsGravitationMCQ+4 / −1
To project a body of mass mmm from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is RE,g=R_E, g=RE​,g= acceleration due to gravity on the surface of earth):
  1. A
    1/2mgRE1 / 2 m g R_E1/2mgRE​
  2. B
    4mgRE4 m g R_E4mgRE​
  3. C
    mgREm g R_EmgRE​
  4. D
    2mgRE2 m g R_E2mgRE​
View written solutionFree

Correct answer: C

  1. Concept used: escape to infinity

To send a body from the Earth's surface to infinity with zero speed at infinity, the minimum required kinetic energy must equal the increase in gravitational potential energy.

  1. Gravitational potential energy at Earth's surface

The gravitational potential energy of a mass mmm at distance RER_ERE​ from the center of Earth is

U=−GMEmREU = -\frac{GM_E m}{R_E}U=−RE​GME​m​

At infinity,

U∞=0U_{\infty} = 0U∞​=0

So the required increase in potential energy is

ΔU=0−(−GMEmRE)=GMEmRE\Delta U = 0 - \left(-\frac{GM_E m}{R_E}\right) = \frac{GM_E m}{R_E}ΔU=0−(−RE​GME​m​)=RE​GME​m​

Hence minimum kinetic energy needed is

K=GMEmREK = \frac{GM_E m}{R_E}K=RE​GME​m​

  1. Use relation between ggg and Earth's parameters

We know

g=GMERE2g = \frac{GM_E}{R_E^2}g=RE2​GME​​

So,

GME=gRE2GM_E = g R_E^2GME​=gRE2​

Substitute into the kinetic energy expression:

K=(gRE2)mRE=mgREK = \frac{(gR_E^2)m}{R_E} = mgR_EK=RE​(gRE2​)m​=mgRE​

  1. Match with options

Thus, the required kinetic energy is

mgRE\boxed{mgR_E}mgRE​​

So the correct option is:

C: mgREmgR_EmgRE​

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