Work Done to Raise a Mass to Height Equal to Earth's Radius

Physics · Gravitation · NEET

To lift a mass m from Earth's surface up to a height equal to Earth's radius R, the work done is W = mgR/2. It is NOT mgR, because gravity gets weaker as you go up, so the average pull is less than the surface value. Memory hook: "Go up one whole radius, pay only half of mgR."
EarthRmass m at surfacelift by h = Rm at distance 2R from centreW = U(2R) - U(R)= (-GMm/2R) - (-GMm/R)= GMm / 2RSince GM = gR^2:W = mgR / 2
Raising mass m from the surface (distance R) to a height h = R (distance 2R from Earth's centre). Using the change in gravitational potential energy gives W = GMm/2R = mgR/2, half of the naive mgR because gravity weakens with distance.

Your doubts, answered

Why is the answer mgR/2 and not mgR?

mgh only works when h is very small compared to R, because then g stays almost constant. When h = R, the mass moves far from Earth, and gravity becomes weaker as distance grows. The correct method uses the change in gravitational potential energy. PE at surface = -GMm/R. PE at height R (distance 2R from centre) = -GMm/2R. Work done = final PE - initial PE = (-GMm/2R) - (-GMm/R) = GMm/2R. Since g = GM/R^2, GM = gR^2, so W = gR^2 m / 2R = mgR/2. The 'half' comes from gravity weakening over the trip.

When can I safely use W = mgh instead?

Use W = mgh only when the height h is very small compared to Earth's radius R (about 6400 km). For example, lifting a bag 2 metres or a lift going 100 m up. There g barely changes, so mgh is accurate. Once h becomes a big fraction of R (like h = R, or h = R/2), you must use the potential energy formula W = GMm/R - GMm/(R+h).

What is the general formula for work done to raise a mass to any height h?

W = GMm[1/R - 1/(R+h)]. Writing GM = gR^2, this becomes W = mgR^2 [1/R - 1/(R+h)] = mgRh/(R+h). Check: put h = R and you get W = mgR(R)/(2R) = mgR/2. Put a tiny h (h << R) and R+h is about R, giving W = mgh, matching the simple case. This one formula covers every height.

Is the work done the same as the change in potential energy?

Yes, when the mass is raised slowly (no leftover speed at the top), the work you do against gravity equals the increase in gravitational potential energy: W = U(final) - U(initial). Because gravitational PE is negative and increases (becomes less negative) as you go up, the work done is positive. This is the safe way to solve any 'raise a mass to height h' NEET problem.

⚠️ The NEET trap
Work done to lift mass m to height R = mgR (using W = mgh with h = R).
Work done = mgR/2, because g is not constant over such a large height; gravity weakens, so the real work is only half of mgR.
🧠 If h is a big fraction of R, mgh is a TRAP. Switch to W = GMm[1/R - 1/(R+h)] = mgRh/(R+h).

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Frequently asked

What is the work done to raise a mass to height equal to Earth's radius?

W = mgR/2, where m is the mass, g is surface gravity, and R is Earth's radius. Equivalently W = GMm/2R.

Why not use mgh for h = R?

Because mgh assumes constant g. At h = R the mass is far from Earth and gravity is weaker, so you must use the potential energy method, giving mgR/2.

What is the work done to raise a mass to height h = R/2?

Use W = mgRh/(R+h). With h = R/2, W = mgR(R/2)/(3R/2) = mgR/3.

What is the general formula for work done to lift a mass to height h?

W = GMm[1/R - 1/(R+h)], which simplifies to W = mgRh/(R+h). It reduces to mgh for small h and mgR/2 for h = R.

Is this work stored as potential energy?

Yes. If the mass is raised slowly, all the work you do against gravity is stored as extra gravitational potential energy.