Gravitational Potential Energy

How to Calculate Gravitational Potential Energy ?

The potential energy of the test particle of mass m located at the point P can be calculated as

$ \displaystyle U(P) = \int_{\infty}^{P} \vec{F}.\vec{dr} $

$ \displaystyle U(P) = GMm\int_{\infty}^{r} \frac{dr}{r^2} $

$ \displaystyle U(P) = -\frac{GMm}{r} $

For the gravitational field due to an extended object,

$ \displaystyle U(P) = \int_{\infty}^{P}( \vec{F_1} +\vec{F_2} +\vec{F_3 }+ … ).\vec{dr} $

Where $\vec{F_1}$  , $\vec{F_2}$ . . . are the forces acting on the test particle due to each point particle within the extended object.

U(P) = U1(P) + U2(P) + . . .

where U1(P) , U2(P) , . . . etc, are the potential energies due to individual particles constituting the object.
This is the result of the application of the superposition principle

Illustration : What is the work done in shifting a body of mass m = 50 kg slowly from the earth’s surface to infinity?

Solution: W = ΔU = Uf – Ui

$ \displaystyle = 0 – (-\frac{GMm}{R} ) $

$ \displaystyle = \frac{GMm}{R} = m g_0 R $

= 3.136 × 109 J

Illustration : What is the gravitational potential energy of the earth and a satellite of mass m = 100 kg orbiting close to the earth’s surface?

Solution: The potential energy between two masses m1 and m2 separated by a distance r is given as:

$ \displaystyle U = -\frac{G m_1 m_2}{r} $

Here,

m1 = mass of earth = M

m2 = mass of the satellite = 100 kg

r = distance between the centre of earth and the satellite ∼ radius of the earth

= R , since the satellite revolves close to earth’s surface.

$ \displaystyle U = -\frac{G M m}{R} $

$ \displaystyle U = -\frac{g R^2 m}{R} = – m g R $

U = – (100 kg) (9.8 m/sec2) (6.4 × 106 m)

= 6.27 × 109 J/kg

Also Read :

∗ Newton’s Law of Gravitation & Gravitational Field Intensity
∗ Acceleration due to Gravity & Variation of g
∗ Work Done in Gravitational Field
∗ Gravitational Potential
∗ Escape Velocity
∗ Kepler’s Laws of Planetary Motion
∗ Motion of Satellites

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