# In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with...

## Question:

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.00 x {eq}10^{16} {/eq} kg and a radius of 8.50 km.

a) What is the speed of a satellite orbiting 4.70 km above the surface?

b) What is the escape speed from the asteroid?

## Orbital Velocity of Satellite:

Just as various planets revolve around the sun, in the same way few celestial bodies revolve around these planets. These bodies are called satellites.

When a satellite revolves in a circular orbit around the earth, a centripetal force acts upon the satellite. This force is the gravitational force exerted by the earth on the satellite.

A satellite of mass m is revolving around the earth with a velocity {eq}v_{0} {/eq} in a circular orbit of radius r. The centripetal force on the satellite is {eq}m\ v_{0}/r {/eq}.

a) The speed of the orbiting satellite is given by:

{eq}\begin{align} v_{0} &= \sqrt{\dfrac{GM}{R + h}}\\ &= \sqrt{\dfrac{(6.67 \times 10^{-11}\ N . m^2/kg^2)(1.00 \times 10^{16}\ kg}{(8.50 \times 10^{3}\ m) + (4.70 \times 10^{3}\ m) }}\\ &= 4.88\ m/s \end{align} {/eq}

Therefore, the speed of the satellite which orbits the asteroid is {eq}4.88\ m/s {/eq}

b) The escape speed in the asteroid is,

{eq}\begin{align} v_{e} &= \dfrac{2GM}{R}\\ &= \dfrac{2(6.67 \times 10^{-11}\ N . m^2/kg^2)(1.00 \times 10^{16}\ kg)}{8.50 \times 10^{3}\ m}\\ &= 8.71\ m/s \end{align} {/eq}

The escape speed on the surface of the asteroid is {eq}8.71\ m/s {/eq} 