Question

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a radius of 8.20 km . You may want to review (Pages 365 - 370) . For help with math skills, you may want to review: Mathematical Expressions Involving Squares For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Phobos escape velocity. Part A Part complete What is the speed of a satellite orbiting 6.00 km above the surface? Express your answer with the appropriate units. 8.11 ms Previous Answers Correct Part B What is the escape speed from the asteroid? Express your answer with the appropriate units.

Answer 1

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

`m(v^2)/((R+h))=(GMm)/((R+h)^2)`

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

`v=\sqrt{(GM)/(R+h)}`

where:

`G=6.67\cdot 10^(-11) m^3 kg^(-1)s^(-2)`

`M=1.40\cdot 10^(16)kg`

`R=8.20 km=8200 m`

`h=6.00 km = 6000 m`

Substituting into the formula,

`v=\sqrt{((6.67\cdot 10^(-11))(1.40\cdot 10^(16)kg))/(8200 m+6000 m)}=8.11 m/s`

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

`v=\sqrt{(2GM)/(R+h)}`

where:

`G=6.67\cdot 10^(-11) m^3 kg^(-1)s^(-2)`

`M=1.40\cdot 10^(16)kg`

`R=8.20 km=8200 m`

`h=6.00 km = 6000 m`

Substituting into the formula, we find:

`v=\sqrt{(2(6.67\cdot 10^(-11))(1.40\cdot 10^(16)kg))/(8200 m+6000 m)}=11.47 m/s`