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Physics Question on Acceleration due to gravity of the earth

Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula –G Mm 1r21r1\frac{1}{r_2 }– \frac{1}{r_1} is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.

Answer

(a) Decreases,
Acceleration due to gravity at depth h is given by the relation: gh ((12hRe)g(1- \frac{2h}{Re} ) g
Where,
Re= Radius of the Earth
g = Acceleration due to gravity on the surface of the Earth

It is clear from the given relation that acceleration due to gravity decreases with an increase in height.


**(b) **Decreases,
Acceleration due to gravity at depth d is given by the relation:
gd (1dRe)g(1-\frac{d}{R_e})g

It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.


**(c) **Mass of the body,
Acceleration due to gravity of body of mass m is given by the relation:
g=GMR2g =\frac{GM }{R_2}
Where,

G = Universal gravitational constant
M = Mass of the Earth
R = Radius of the Earth

Hence, it can be inferred that acceleration due to gravity is independent of the mass body.


**(d) **More,
Gravitational potential energy of two points Earth is respectively given by:

V(r1) = -GmMr1\frac{GmM }{r_1}
V(r2) = - GmMr2\frac{GmM }{r_2}
∴ Difference in potential energy, V=V(r2)V(r1)=GmM(1r21r1)V = V(r_2) - V(r_1) = -GmM (\frac{1}{r_2 }- \frac{1}{r_1})

Hence, this formula is more accurate than the formula mg(r2– r1).