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"g = 9.8 m/s²" is a good number for the classroom, but it is not the same everywhere. It is smaller on a mountain, smaller down a mine, and smaller at the equator than at the poles. JEE questions test all three effects, and the common error is using a shortcut where it does not apply. This post derives each formula, works the numbers, and shows how the real Earth differs from the exam model.
Which syllabus covers this?
The JEE (Main) 2026 syllabus lists "acceleration due to gravity and its variation with altitude and depth" in Unit 6, and NCERT's Gravitation chapter derives both in section 7.6. Neither NCERT's chapter nor the JEE Main syllabus mentions the rotation effect. The JEE (Advanced) 2026 brochure lists "acceleration due to gravity" without listing sub-topics, so treat rotation as an extension. Check your exam year's syllabus.
How does g change with height?
Above the surface you are outside all the mass, so g(h) = GM/(R + h)² = g(1 + h/R)⁻². For h much smaller than R, NCERT uses the binomial approximation g(h) ≈ g(1 − 2h/R).
How does g change with depth?
At depth d, NCERT splits the Earth into an inner sphere of radius R − d and an outer shell. The shell exerts no force on a point inside it, and the inner sphere acts as if its mass were at the centre. For uniform density its mass is proportional to (R − d)³, which gives g(d) = g(1 − d/R). For more on the shell result, see our post on gravity inside a shell and a sphere with a cavity.
Worked example: 64 km up and 64 km down
Take R = 6400 km and g = 9.8 m/s².
64 km up, exact: 9.8 × (6400/6464)² ≈ 9.607 m/s².
64 km up, approximate: 9.8 × (1 − 2 × 64/6400) = 9.8 × 0.98 ≈ 9.604 m/s². The shortcut is good here.
64 km down: 9.8 × (1 − 64/6400) = 9.8 × 0.99 ≈ 9.702 m/s².
For small distances, g falls twice as fast going up as going down. A 1% drop needs a climb of 32 km or a descent of 64 km.
For large distances the approximation breaks. At h = R the exact formula gives g/4. The shortcut gives g(1 − 2) = −g, which is nonsense. Under the uniform model, g also falls to g/4 at depth d = 3R/4, which is 4800 km.
Is the real Earth uniform?
No. OpenStax University Physics shows that a uniform planet gives g falling in a straight line to zero at the centre, then notes that Earth's density is not constant and that a model based on seismic data (PREM) is probably closer to the real profile. The formula g(1 − d/R) is an exam model.
How does Earth's rotation change g?
A body at the equator moves in a circle of radius R once a day, so part of gravity's pull is used up providing the centripetal acceleration Rω². OpenStax works this out as about 0.0337 m/s², or 0.34% of g.
Using NASA's Earth fact sheet values of a sidereal rotation period of 23.9345 hours and an equatorial radius of 6378.137 km: ω = 2π/(23.9345 × 3600) ≈ 7.29 × 10⁻⁵ rad/s, and Rω² ≈ 0.0339 m/s².
At latitude λ the circle has radius R cos λ, and to a very good approximation the effective value is g' = g − Rω²cos²λ. At 30°, cos²λ = 0.75, so the reduction is about 0.025 m/s². At the poles it is zero.
NASA's fact sheet lists surface acceleration of 9.780 m/s² at the equator and 9.832 m/s² at the pole, a difference of 0.052 m/s². Rotation explains only part of that. Earth also bulges: NASA gives an equatorial radius of 6378.137 km against a polar radius of 6356.752 km, and OpenStax notes that the shape effect is comparable to the rotation effect and acts in the same direction.
OpenStax also asks how fast Earth would need to spin for people at the equator to feel weightless. The answer is a day of about 84 minutes.
Which mistakes cost marks?
Using g(1 − 2h/R) for heights comparable to R.
Mixing up the factor 2: height has 2h/R, depth has d/R.
Treating the rotation correction as the same at every latitude.
Assuming g is largest at the centre of the Earth. In the uniform model it is zero there.
A practice routine
Derive g(h) and g(d) from scratch, including the shell argument.
Sketch g against distance from the centre, from 0 to 3R, and mark R.
Find the height at which g equals its value at a depth of R/2. (Answer: (√2 − 1)R, about 0.41R.)
Compute the rotation correction at your own city's latitude.
For a step you cannot picture, such as why the outer shell exerts no force, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson. Our post on gravitational potential energy beyond mgh uses the same idea of g changing with distance.
Tonight, draw the g-versus-r graph from memory and check its shape against the formulas here.
