Back to Course

SS3: PHYSICS - 1ST TERM

0% Complete
0/0 Steps



  • Do you like this content?

  • Follow us

Lesson 3, Topic 4
In Progress

Gravitational Potential (V)

Lesson Progress
0% Complete

Whenever a body is lifted against the force of gravity, 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 energy. 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 vr 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

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

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

= 6.523 x 10 10

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

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

= -5.408 x 10 10 J

Gain in P.e = Vr – Vo

= -5.408 x 10 10 – (-6.523 x 10 10)

=8.45 10 9J

Responses

Your email address will not be published. Required fields are marked *

back-to-top
error: