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Q. 32

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 354

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Short Answer

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

See the step by step solution

Step by Step Solution

Step 1 : Given information a.

Height to be calculated for free-fall acceleration to be of its value at the surface.

Step 2 : Calculation a.

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.

Here,

Step 3 : Final answer a.

At a height of above the earth the free-fall acceleration is of its value at the surface.

Step 4 : Given information b.

The height of the satellite is

Step 5 : Calculation b.

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.

Now,

Where orbital radius of satellite : r = R + h

And

Here,

Radius of earth

Now comparing we get,

Substituting values and solving we get,

Step 6 : Final answer b.

The speed of the satellite is

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