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JEE Physics 2023 Question with Solution

A space ship of mass 2×104 kg2 \times 10^4 \, \text{kg} is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if g=10 m/s2g = 10 \, \text{m/s}^2 and radius of earth = 6400 km6400 \, \text{km}):

  • A

    7.9(2−1) km/s7.9(\sqrt{2}-1) \, \text{km/s}

  • B

    7.4(2−1) km/s7.4(\sqrt{2}-1) \, \text{km/s}

  • C

    11.2(2−1) km/s11.2(\sqrt{2}-1) \, \text{km/s}

  • D

    8(2−1) km/s8(\sqrt{2}-1) \, \text{km/s}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A space ship is in a circular orbit close to the earth surface, with g=10 m/s2g = 10 \, \text{m/s}^2 and radius of earth R=6400 km=6.4×106 mR = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m}.

Find: The additional velocity needed to escape from orbit.

For a body in a circular orbit near earth, the orbital speed is

vo=gRv_o = \sqrt{gR}

and the escape speed from the same point is

ve=2gRv_e = \sqrt{2gR}

So the additional speed required is

ΔV=ve−vo=gR(2−1)\Delta V = v_e - v_o = \sqrt{gR}(\sqrt{2}-1)

Step-by-step Evaluation

Using g=10 m/s2g = 10 \, \text{m/s}^2 and R=6.4×106 mR = 6.4 \times 10^6 \, \text{m},

gR=10×6.4×106=6.4×107gR = 10 \times 6.4 \times 10^6 = 6.4 \times 10^7

Hence,

gR=6.4×107=8×103 m/s=8 km/s\sqrt{gR} = \sqrt{6.4 \times 10^7} = 8 \times 10^3 \, \text{m/s} = 8 \, \text{km/s}

Therefore,

ΔV=8(2−1) km/s\Delta V = 8(\sqrt{2}-1) \, \text{km/s}

So the correct option is D.

The solution labels option C, but the worked value matches option D exactly.

Common mistakes

  • Using escape speed directly as the required additional speed. This is wrong because the space ship is already moving in orbit with orbital speed. Subtract the orbital speed from the escape speed instead.

  • Forgetting to convert the earth radius from 6400 km6400 \, \text{km} to metres when using gg in SI units. Use consistent units before substitution.

  • Using the mass of the space ship in the calculation. This is unnecessary because both orbital speed and escape speed are independent of the spacecraft mass.

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