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JEE Physics 2024 Question with Solution

A satellite of 103 kg10^3 \, \text{kg} mass is revolving in a circular orbit of radius 2R2R. If 104R6\frac{10^4R}{6} joules of energy is supplied to the satellite, it would revolve in a new circular orbit of radius:

  • A

    2.5R2.5R

  • B

    3R3R

  • C

    4R4R

  • D

    6R6R

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mass of satellite = 103 kg10^3 \, \text{kg}, initial orbit radius = 2R2R, energy supplied = 104R6\frac{10^4R}{6} joules.

Find: The new circular orbit radius.

For a satellite in a circular orbit, the total mechanical energy is

E=−GMm2rE = -\frac{GMm}{2r}

Initially, for orbit radius 2R2R,

Ei=−GMm2(2R)=−GMm4RE_i = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}

After supplying energy 104R6\frac{10^4R}{6}, the new total energy becomes

−GMm4R+104R6=−GMm2r-\frac{GMm}{4R} + \frac{10^4R}{6} = -\frac{GMm}{2r}

where rr is the new radius. Using

g=GMR2g = \frac{GM}{R^2}

we get

−mgR4+104R6=−mgR22r-\frac{mgR}{4} + \frac{10^4R}{6} = -\frac{mgR^2}{2r}

Substituting m=103 kgm = 10^3 \, \text{kg} and g=10 m/s2g = 10 \, \text{m/s}^2,

−103×10×R4+104R6=−103×10×R22r-\frac{10^3 \times 10 \times R}{4} + \frac{10^4R}{6} = -\frac{10^3 \times 10 \times R^2}{2r}

Dividing throughout by 104R10^4R,

−14+16=−R2r-\frac{1}{4} + \frac{1}{6} = -\frac{R}{2r} −112=−R2r-\frac{1}{12} = -\frac{R}{2r} r=6Rr = 6R

Therefore, the satellite will revolve in the new circular orbit of radius 6R6R. The correct option is D.

Energy Change Interpretation

Given: The satellite moves from an initial circular orbit of radius 2R2R to a new circular orbit after addition of energy 104R6\frac{10^4R}{6}.

Find: The new orbit radius.

The total mechanical energy of a satellite in a circular orbit is

E=−GMm2rE = -\frac{GMm}{2r}

When energy is supplied, the total energy increases, so the orbit radius also increases.

Let the new radius be r′r'. Then

Einitial+Energy supplied=EfinalE_{\text{initial}} + \text{Energy supplied} = E_{\text{final}} −GMm2(2R)+104R6=−GMm2r′-\frac{GMm}{2(2R)} + \frac{10^4R}{6} = -\frac{GMm}{2r'}

On simplifying this relation using the values used in the solution, we obtain

r′=6Rr' = 6R

Hence, the new circular orbit radius is 6R6R, so the correct option is D.

Common mistakes

  • Using kinetic energy alone instead of total mechanical energy is incorrect because orbital radius depends on the total energy E=−GMm2rE = -\frac{GMm}{2r}. Always use total mechanical energy for circular satellite orbits.

  • Forgetting that the initial radius is 2R2R leads to a wrong initial energy expression. The correct initial energy is −GMm4R-\frac{GMm}{4R}, not −GMm2R-\frac{GMm}{2R}.

  • Substituting GM=gRGM = gR is wrong. Since g=GMR2g = \frac{GM}{R^2}, the correct substitution is GM=gR2GM = gR^2.

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