Set 51 Problem number 3

Problem

A charge of 32 `microC, located at the origin, is attached to a large mass and can be considered stationary.  A charge of -44.01 `microC is attached to a relatively small mass of .0032 kg. The second charge orbits the first in a circular orbit which passes through ( .42 m, 0, 0). What is its orbital velocity?

Solution

The key here is that the Coulomb force of attraction must provide the centripetal force which results in the centripetal acceleration v^2/r of the orbiting mass.

• The force of attraction has magnitude | (9 x 10^9 N m^2/C^2)( 32 x 10^-6 C)(-44.01 `micro C)/( .42 m)^2 | = 71.85 N; this is the centripetal force.
• The centripetal acceleration is thus F / m = 71.85 N / ( .0032 kg) = 22450 m/s^2.
• This centripetal force is equal to v^2/r, where r is the radius of the orbit--in this case the .42 m separation of the charges. Thus we have v^2/r = v^2/( .42 m) = 22450 m/s^2.
• Solving, we get v = `sqrt[ 22450 m/s^2 ( .42 m)] = 97.1 m/s.

Generalized Solution

An orbit of radius r and velocity v requires a centripetal acceleration of a = v^2 / r. If the orbiting mass is m, then the corresponding centripetal force required to maintain the circular path is F = ma = m v^2 / r. In this case the centripetal force is supplied by the Coulomb attraction | F | = k | q1 | | q2 | / r^2 . Thus we have

m v^2 / r = k | q1 | | q2 | / r^2.

We easily solve this equation for v to obtain

v = `sqrt( k | q1 q2 | / (m r) ).

Explanation in terms of Figure(s), Extension

The figure below shows the Coulomb force F on the orbiting charge. The circular orbit bounds the shaded circle. The centripetal acceleration a is directed toward the center of the circle, where the fixed charge q1 resides.