a. At what height above the earth is the free-fall acceleration 10%of its value at the surface?

b. What is the speed of a satellite orbiting at that height?

Short Answer

Expert verified

a. At a height of 1.38×107mabove the earth is the free-fall acceleration 10%of its value at the surface.

b. The speed of the satellite orbiting at that height is4.44×103m/s.

Step by step solution

01

Given information a.

Height to be calculated for free-fall acceleration to be 10%of its value at the surface.

02

Calculation a.

Acceleration due to gravity at height h above the surface is given by : gh=gR2(R+h)2

Where, g = value at surface.

gh= value at height h.

R = radius of earth.

h = height above the surface of earth.

Here,

gh=10%ofg,sogh=0.1gR=6.37×106m

03

Final answer a.

At a height of 1.38×107mabove the earth the free-fall acceleration is10%of its value at the surface.

04

Given information b.

The height of the satellite is1.38×107m.

05

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, Fc=centripetal force=mv2r

Where orbital radius of satellite : r = R + h

And Fg=GMmr2

Here,

h=1.38×107mM=mass of earth=5.97×1024kgG=6.67×10-11Nm2kg2

Radius of earthR=6.371×106m

Now comparing we get,

Fc=Fgmv2r=GMmr2v=GMr=GMR+h

Substituting values and solving we get,v=4.44×103m/s.

06

Final answer b.

The speed of the satellite is4.44×103m/s.

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