What is the relation between height 'h' and depth 'd' for the same change in 'g'?
Answers
At a height of h from the surface of the earth, the gravitational force on an object of mass m is
F = GMm/(R+h)^2
Here (R + h) is the distance between the object and the centre of earth.
Say at that height h, the gravitational acceleration is g1.
So we can write, mg1 = GMm / (R+h)^2
=> g1 = GM/(R+h)^2 _________________ (5)
Now we know on the surface of earth, it is
g = GM / R^2
Taking ratio of these 2,
g1/g = R^2 /(R+h)^2
= 1/(1 + h/R)^2 = (1 + h/R)^(-2) = (1 – 2h/R)
=> g1/g = (1 – 2h/R)
=> g1 = g (1 – 2h/R) ___________________ (6)
So as altitude h increases, the value of acceleration due to gravity falls.
Let’s say, a body of mass m is resting at point A , where A is at a depth of h from the earth’s surface.
Distance of point A from the centre of the earth = R – h,
where R is the radius of the earth.
Mass of inner sphere = (4/3). Pi. (R-h)^3. p
Here p is the density.
Now at point A, the gravitational force on the object of mass m is
F = G M m/ (R-h)^2 = G. [(4/3). Pi. (R-h)^3. p] m/(R-h)^2 = G. (4/3). Pi. (R-h). p. m
Again at point A, the acceleration due to gravity (say g2) = F/m = G. (4/3). Pi. (R-h). p _________________ (7)
Now we know at earth’s surface, acceleration due to gravity = g = (4/3) Pi R p G
Taking the ratio, again,
g2/g = [G. (4/3). Pi. (R-h). p ]/ [(4/3) Pi R p G] = (R-h) / R = 1 – h/R.
=> g2 = g (1 – h/R) _________________________ (8)
So as depth h increases, the value of acceleration due to gravity falls.
HOPE IT HELPS YOU
Answer:2h= d
where h is height and d is depth
Explanation: