StemCET Logo

Physics Question 32 – JEE-MAIN 2026

A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth (Re). The increase in potential energy will be _______. (g is acceleration due to gravity at the surface of earth)

Recall the formula for gravitational potential energy of a mass m at a distance r from the center of a planet of mass M.

Step 1: Identify Initial and Final Positions✦ Active

The body starts at the Earth's surface, so its initial distance from the Earth's center is r1=Re. It is moved to a height h=2Re from the surface, so its final distance from the Earth's center is r2=Re+h=Re+2Re=3Re.

Step 2: Calculate Potential Energy at Each Position○ Expand

The gravitational potential energy of a mass m at a distance r from the center of Earth (mass M) is given by U=GMmr.

U1=GMmRe
U2=GMm3Re
💡 Teacher's Secret Hint

Remember that gravitational potential energy is negative and increases (becomes less negative) as distance from the center increases.

Step 3: Determine the Increase in Potential Energy○ Expand

The increase in potential energy is ΔU=U2U1. We also know that the acceleration due to gravity at the surface is g=GMRe2, which implies GM=gRe2.

ΔU=GMm3Re(GMmRe)=GMmReGMm3Re
ΔU=GMm(1Re13Re)=GMm(313Re)=2GMm3Re

Substitute GM=gRe2 into the expression for ΔU:

ΔU=2(gRe2)m3Re=2mgRe23Re=23mgRe
💡 Teacher's Secret Hint

Ensure correct algebraic manipulation of fractions and substitution of GM in terms of g and Re.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.