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MCQMediumJEE 2023Acceleration due to Gravity

JEE Physics 2023 Question with Solution

A planet has double the mass of the Earth. Its average density is equal to that of the Earth. An object weighing WW on Earth will weigh on that planet:

  • A

    21/3W2^{1/3}W

  • B

    2W2W

  • C

    WW

  • D

    22/3W2^{2/3}W

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The planet has mass Mp=2MeM_p = 2M_e and average density equal to Earth, so ρp=ρe\rho_p = \rho_e.

Find: The weight of an object on the planet in terms of WW.

Using density,

ρ=M43πR3\rho = \frac{M}{\frac{4}{3}\pi R^3}

Since ρp=ρe\rho_p = \rho_e,

MeRe3=MpRp3\frac{M_e}{R_e^3} = \frac{M_p}{R_p^3}

Substituting Mp=2MeM_p = 2M_e,

MeRe3=2MeRp3\frac{M_e}{R_e^3} = \frac{2M_e}{R_p^3}

So,

Rp3=2Re3R_p^3 = 2R_e^3

Hence,

Rp=21/3ReR_p = 2^{1/3} R_e

The acceleration due to gravity is

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

Therefore,

gpge=Mp/Rp2Me/Re2\frac{g_p}{g_e} = \frac{M_p/R_p^2}{M_e/R_e^2}

Substituting Mp=2MeM_p = 2M_e and Rp=21/3ReR_p = 2^{1/3}R_e,

gpge=2Me(21/3Re)2⋅Re2Me=222/3=21/3\frac{g_p}{g_e} = \frac{2M_e}{\left(2^{1/3}R_e\right)^2} \cdot \frac{R_e^2}{M_e} = \frac{2}{2^{2/3}} = 2^{1/3}

Thus,

gp=21/3geg_p = 2^{1/3} g_e

Weight is W=mgW = mg. Since the mass of the object remains the same,

Wp=mgp=m(21/3ge)=21/3mge=21/3WW_p = m g_p = m\left(2^{1/3} g_e\right) = 2^{1/3} m g_e = 2^{1/3} W

Therefore, the object will weigh 21/3W2^{1/3}W on the planet. The correct option is A.

Using proportionality of gravity with density and radius

Given: The planet has double Earth’s mass and the same average density.

Find: Weight on the planet in terms of WW.

For a spherical planet of uniform average density,

M∝R3M \propto R^3

when density is constant. Since the new planet has mass 2Me2M_e, its radius becomes

Rp∝(2Me)1/3R_p \propto (2M_e)^{1/3}

So,

Rp=21/3ReR_p = 2^{1/3} R_e
Derivation showing ratio of gravitational accelerations using mass, radius, and density, concluding the factor becomes two raised to power one by three.

Now,

g∝MR2g \propto \frac{M}{R^2}

Hence,

gp∝2Me(21/3Re)2=21/3MeRe2g_p \propto \frac{2M_e}{\left(2^{1/3}R_e\right)^2} = 2^{1/3} \frac{M_e}{R_e^2}

Therefore,

gp=21/3geg_p = 2^{1/3} g_e

and so the weight becomes

Wp=21/3WW_p = 2^{1/3} W

Therefore, the correct answer is 21/3W2^{1/3}W.

Common mistakes

  • Using g∝Mg \propto M only and ignoring the change in radius. This is wrong because gravity depends on both mass and radius as g=GMR2g = \frac{GM}{R^2}. First find how the radius changes from the density condition.

  • Assuming the radius also doubles when the mass doubles. This is wrong because equal density implies M∝R3M \propto R^3, so the radius scales as R∝M1/3R \propto M^{1/3}, not directly as mass.

  • Confusing mass of the object with its weight. The object’s mass remains constant, while its weight changes because gg changes. Compute the new gg first, then multiply by the same object mass.

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