Acceleration due to gravity and its variation with altitude and depth
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See all →- A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is : Answer: 2. 27 N Key idea: weight at height h varies as 1/(R+h)² . F=48(R/R+R/3)²=48×(3/4)²=48×frac916=27\ mathrm N 1
- The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is : Answer: 1. 3.92 m s ⁻² Key idea: g=GM/R² . For the planet M_p=M/10 and R_p=R/2 , so g_p=G(M/10)/(R/2)²=4/10 g=0.4×9.8=3.92\ m s⁻² Option 1. 1
- A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ? Answer: 2. 32 N Gravitational force varies as g , and g at height h is g(R/R+h)² . With h=R/2 : (R/1.5R)²=4/9 . F=72×4/9=32\ N Answer: optio... 1
- A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth ? Answer: 4. 100 N Inside the earth at depth d , g'=g(1-d/R) . Halfway to the centre means d=R/2 , so g'=g/2 . Weight =200/2=100 N. 1
- The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then : Answer: 3. d = 2 km Key idea: at height h (small): g'=g(1-2h/R) . At depth d : g'=g(1-d/R) . Equating: 2h/R=d/R⇒ d=2h=2 km. 1
- Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by Answer: 1. graph (1) Inside the earth g is proportional to the distance from the centre; outside it falls as 1/r² . For r For r>R : g=GM/r²... 1