Gravitational Potential (V)

Whenever a body is lifted against the force of gravityThe force of gravity, or gravitational force, pulls objects with mass toward each other. More, work is done on the body. Likewise, when a body is lifted from earth’s surface to a new position above the earth’s surface, the work is stored in the body as potential energyEnergy is the ability to do work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, and electrical energy. Units of Energy: The SI unit... More. E.g. if a body is moved from a point to another point r from the earth’s surface, the work done = force x displacement

= \( \frac{GMM} {r^2} \scriptsize \; \times \; r\)

= \( \frac{GMM} {r}\)

V = \( \frac{GMM} {r}\)

Where M- Mass of the earth, m-mass of body moved against the earth

r- Distance of the mass from earth’s surface 

A man outside the earth’s gravitational field is at infinity ( ∞) and the gravitational potential energy is zero at infinity.

When a mass at infinity is moved to a new location on the earth’s surface, work is done, and v = \( \scriptsize v_ {\infty} – v_r \)

 \( \scriptsize v_r \) = potential at new location

\( \scriptsize v_{\infty} \) =potential at infinity

\( \frac{GMM} {r} \scriptsize = 0 – v_r \) \( \scriptsize v_r = \normalsize \frac{GMM} {r} \)

Work done per unit mass is called gravitational potential, = \( \frac{GM} {r}\)

Therefore, gravitational potential is the work done in moving a unit mass (kg) from infinity to other point in the earth’s gravitational field.

The negative sign (-) shows that potential at v_r is less than potential at infinity, \( \scriptsize v_{\infty} \)

A rocket of mass 1000kg is resting on the earth’s surface. If the earth’s radius is 6.4 x 10 ⁶  and mass is 6.0 x 10 ²⁴, calculate:

• P.e of the rocket on earth’s surface
• P.e of the rocket at height 1.0  x 10 ⁶ m
• Gain in P.e of the rocket position in the new position

V_y = \( \frac{GMm} {r}\)

= \( \frac{6.67 \times 10^{-11} \; \times \; 1000 \; \times \; 6.0 \times 10^{24}} {6.4 \times 10^{6}}\)

= 6.523 x 10 ¹⁰

V_r = \( \scriptsize – v_0 \normalsize \frac{r}{r+h} \) 

= \( \scriptsize – 6.523 \times 10^{10} \; \times \; \normalsize \frac{6.4 \times 10^{6}}{6.0 \times 10^{-24} + 1.0 \times 10^{6}} \) 

= -5.408 x 10 ¹⁰ J

Gain in P.e = V_r – V_o

= -5.408 x 10 ¹⁰ – (-6.523 x 10 ¹⁰)

=8.45 10 ⁹J