Physics · Gravitation · NEET
g comes from the formula g = GM/r^2, where r is the distance from the Earth's centre. When you rise to a height h, your distance becomes r = R + h, which is larger. Since g depends on 1/r^2, a bigger r gives a smaller g. So g_h = GM/(R+h)^2 = g/(1 + h/R)^2. This is why a body weighs less on a high mountain or in a plane than at sea level.
Use it only when h is very small compared to R (for example, a mountain or a low-flying plane, h a few km vs R = 6400 km). It comes from expanding (1 + h/R)^-2 by the binomial theorem and keeping only the first term: (1 + h/R)^-2 is about 1 - 2h/R for small h/R. If h is large (like h = R or a satellite far up), this short form is wrong — you must use the exact g_h = g/(1 + h/R)^2.
No. For height, the approximate formula has a 2: g_h = g(1 - 2h/R). For depth, there is no 2: g_d = g(1 - d/R). So g falls off twice as fast at height as it does at the same depth. That is why in the NEET 2017 question, equal g at height h and depth d gives d = 2h. Do not mix them up — this exact swap is a favourite NTA trap.
No. Mass (in kg) is the amount of matter and stays constant everywhere. What changes is weight, W = mg. As g decreases with height, the weight decreases, but the mass is unchanged. So a 5 kg bag is still 5 kg on a mountain, but it presses down with slightly less force.
Put h = R into the exact formula: g_h = g/(1 + R/R)^2 = g/(1 + 1)^2 = g/4. So at h = R, g is only one-quarter of its surface value, about 2.45 m/s^2. Do not use the short 1 - 2h/R form here — h = R is not small, and it would wrongly give a negative answer.
A body weighs 48 N on the surface of the earth. The gravitational force on it at a height equal to one-third the radius of the earth from the surface is:
The acceleration due to gravity at a height 1 km above the earth equals that at a depth d below the surface. Then:
At what height from the surface of earth the gravitational potential and the value of g are -5.4x10^7 J/kg and 6.0 m/s^2 respectively? (Radius of earth = 6400 km)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The exact formula is g_h = g/(1 + h/R)^2, where g is surface gravity, R is Earth's radius, and h is the height. For small heights (h << R) you may use the approximation g_h = g(1 - 2h/R).
g decreases with height. As you rise, your distance from the Earth's centre (R + h) increases, and since g depends on 1/distance^2, g becomes smaller.
Set g/(1 + h/R)^2 = g/2, so (1 + h/R)^2 = 2, giving 1 + h/R = sqrt(2). Then h = R(sqrt(2) - 1) = about 0.414 R, roughly 2650 km.
The 2 comes from the binomial expansion of (1 + h/R)^-2 for small h/R, which gives 1 - 2h/R. The depth formula comes from a different physics (only the inner sphere pulls you) and gives 1 - d/R, with no 2. So g falls twice as fast with height as with depth.
Only weight changes. Mass is constant everywhere. Weight W = mg decreases with height because g decreases, but the number of kilograms stays the same.
