a. At what height above the earth is the free-fall acceleration of its value at the surface?
b. What is the speed of a satellite orbiting at that height?
a. At a height of above the earth is the free-fall acceleration of its value at the surface.
b. The speed of the satellite orbiting at that height is
Height to be calculated for free-fall acceleration to be of its value at the surface.
Acceleration due to gravity at height h above the surface is given by :
Where, g = value at surface.
= value at height h.
R = radius of earth.
h = height above the surface of earth.
At a height of above the earth the free-fall acceleration is of its value at the surface.
The height of the satellite is
For a satellite to revolve around the planet the necessary centripetal force is provided by the gravitational force of attraction between the planet and the satellite.
Where orbital radius of satellite : r = R + h
Radius of earth
Now comparing we get,
Substituting values and solving we get,
The speed of the satellite is
The mass of Jupiter is 300 times the mass of the earth. Jupiter orbits the sun with TJupiter = 11.9 yr in an orbit with rJupiter = 5.2rearth. Suppose the earth could be moved to the distance of Jupiter and placed in a circular orbit around the sun. Which of the following describes the earth’s new period? Explain. a. 1 yr b. Between 1 yr and 11.9 yr c. 11.9 yr d. More than 11.9 yr e. It would depend on the earth’s speed. f. It’s impossible for a planet of earth’s mass to orbit at the distance of Jupiter.
94% of StudySmarter users get better grades.Sign up for free