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 Gravitational Potential Energy and Gravitational Potential
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Gravitational Potential Energy

Potential energy is due to the virtue of position of the object.

Gravitational Potential Energy is due to the potential energy of a body arising out of the force of gravity.

Consider a particle which is at a point P above the surface of earth and when it falls on the surface of earth at position Q, the particle is changing its position because of force of gravity.

The change in potential energy from position P to Q is same as the work done by the gravity.

It depends on the height above the ground and mass of the body.
Expression for gravitational potential energy:
Case 1: 'g' is Constant.
Consider an object of mass 'm' at point A on the surface of the earth.
work done will be given as;
`"W"_("BA")="FX"` Displacement where F=gravitational force exerted towards the earth.
=mg(h_{2}h_{1}) (body is brought from position A to B)
=mgh_{2}  mgh_{1}
W_{AB = }V_{A  }V_{B }
Where,
V_{A} = potential energy at point A
V_{B} = Potential energy at point B
From above equation we can say that the work done in moving the particle is just the difference of potential energy between its final and initial positions.
Case 2: 'g' is not constant.
Calculate work done in lifting a particle from r = r_{1} to r = r_{2}(r_{2} > r_{1}) along a vertical path .
we will get, W=V(r_{2})  V(r_{1})
Conclusion:
In general the gravitational potential energy at a distance 'r' is given by `"V"(r)=("GM"_em)/"r" + "V"_o`
where,
V(r)= potential energy at a distance 'r'.
V_{o} = At this point gravitational potential energy is zero.
Gravitational potential energy is `prop` to the mass of the particle.
Problem:
Choose the correct alternative:
 Acceleration due to gravity increases/decreases with increasing altitude.
 Acceleration due to gravity increases/decreases with increasing depth. (assume the earth to be a sphere of uniform density).
 Acceleration due to gravity is independent of mass of the earth/mass of the body.
 The formula – `GMm(1/r_2– 1/r_1)` is more/less accurate than the formula `"mg"(r_2– r_1)` for the difference of potential energy between two points r_{2}and r_{1} distance away from the centre of the earth.
Answer:
(a)Decreases
(b)Decreases
(c)Mass of the body
(d)More
Explanation:
Acceleration due to gravity at depth h is given by the relation:
`"gh" = (1 "2h"/R_E)g`
Where,
R_{E} = 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.
Acceleration due to gravity at depth d is given by the relation:
`"gd"=(1"d"/"R"_E)g`
It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.
Acceleration due to gravity of body of mass m is given by the relation: `"g"="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 of the body.
Gravitational potential energy of two points r_{2} and r_{1} distance away from the centre of the Earth is respectively given by:
`"V"(r_1) =  "GmM"/r_1`
`"V"(r_2) = "GmM"/r_2`
Therefore,
Difference in potential energy, V = V(r_{2}) – V(r1) =GmM (1/r_{2} – 1/r_{1})
Hence, this formula is more accurate than the formula mg (r_{2}– r_{1}).