Question

A large asteroid of mass 33900 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 610 kg, is in a circular orbit about the first at a distance of 146 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid? Now suppose that the first and second asteroids carry charges of 1.18 C and -1.18 C, respectively. How fast would the second asteroid have to be moving in order to occupy the same circular orbit as before?

Answer 1

Answer:

a) 1.2*10^-4 m/s

b) 375 m/s

Explanation:

I assume the large asteroid doesn't move.

The smaller asteroid is affected by an acceleration determined by the universal gravitation law:

a = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s^2))

M: mass of the large asteroid (33900 kg)

d: distance between them (146 m)

Then:

a = 6.67*10^-11 * 33900 / 146^2 = 10^-10 m/s^2

I assume the asteroid in a circular orbit, in this case the centripetal acceleration is:

a = v^2/r

Rearranging:

v^2 = a * r

$v = \sqrt{a * r}$

v = \sqrt{10^-10 * 146} = 1.2*10^-4 m/s

If the asteroids have electric charges of 1.18 C and -1.18 C there will be an electric force of:

F = 1/(4π*e0)*(q1*q2)/d^2

Where e0 is the electrical constant (8.85*10^-12 F/m)

F = 1/(4π*8.85*10^-12) (-1.18*1.18)/ 146^2 = -587 kN

On an asteroid witha mass of 610 kg this force causes an acceleration of:

F = m * a

a = F / m

a = 587000 / 610 = 962 m/s^2

With the electric acceleration, the gravitational one is negligible.

The speed is then:

v = \sqrt{962 * 146} = 375 m/s