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Chapter 5, Problem 71P

(a)

To determine

The acceleration due to gravity at a height 350km above the sea level , the orbital velocity and the time period of the space station.

(a)

Expert Solution
Check Mark

Answer to Problem 71P

The acceleration due to gravity is 8.8m/s2_, the orbital velocity is 7700m/s_ and 5500s_ is the time period.

Explanation of Solution

The gravitational force of an object on the Earth’s surface is determined by Newton’s universal law of gravitation. Which is given by,

    Fgrav=GMEarthMstationrothit2        (I)

Here, Fgrav is the gravitational force, G is the gravitational constant, MEarth is the mass of the earth, Mstation is the mass of the station.

Force of gravity is the only force acting on the station so that for an object in orbit it must be providing the entire centripetal acceleration,

    GMEarthMstationrothit2=MStation v2rorbit v=GMEarthrorbit        (II)

The acceleration of the station when it is moving in a circle must be centripetal acceleration which is given by,

    astation =v2rorbit         (III)

The expression for the time period is given by,

    T=2πrv        (IV)

Conclusion:

The orbit of the station is equal to Earths radius plus height above the surface so that, r=6.37×106m+3.50×105m=6720000

Substitute 6.67×1011Nm2kg for G, 5.97×1024kg for MEarth, 6720000m for r in equation (II) to find v.

    v=(6.67×1011Nm2kg)(5.97×1024kg)(6.37×106m+3.50×105m)=7697m/s7700m/s

Substitute 7697m/s for v and 6720000m for r in equation (III) to find astation.

    astation=(7697m/s)2(6.37×106m+3.50×105m)=8.8m/s2

Substitute 7697m/s for v and 6720000m for r in equation (IV) to find T.

    T=2π(6.37×106m+3.50×105m)7697m/s=5500s

Therefore, The acceleration due to gravity is 8.8m/s2_, the orbital velocity is 7700m/s_ and 5500s_ is the time period.

(b)

To determine

The orbital velocity and period of the telescope.

(b)

Expert Solution
Check Mark

Answer to Problem 71P

The orbital velocity is 7600m/s_ and 5800s_ is the time period.

Explanation of Solution

Substitute 6.67×1011Nm2kg for G, 5.97×1024kg for MEarth, (6.37×106m+6.00×105m) for r in equation (II) to find v.

    v=(6.67×1011Nm2kg)(5.97×1024)kg(6.37×106m+6.00×105m)=7560m/s7600m/s

Conclusion:

Substitute (6.37×106m+6.00×105m) for r and 7560m/s for v in equation (IV) to find T.

    T=2π(6.37×106m+6.00×105m)7560m/s=5800s

Therefore, The orbital velocity is 7600m/s_ and 5800s_ is the time period.

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Chapter 5 Solutions

Bundle: College Physics: Reasoning And Relationships, 2nd + Webassign Printed Access Card For Giordano's College Physics, Volume 1, 2nd Edition, Multi-term

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