1. A satellite is in orbit about Earth. Its orbital radius is 3.76×10 7 m. The m

Question

1. A satellite is in orbit about Earth. Its orbital radius is 3.76×107 m. The mass of the satellite is 4589 kg and the mass of Earth is 5.974×1024 kg. Determine the orbital speed of the satellite in mi/s. 1 mi/s = 1609 m/s.

2. Cloverleaf exits are approximately circular. A 632 kg automobile is traveling 56 mph while taking a cloverleaf exit that has a radius of 40.1 m. Calculate the centripetal acceleration of the car and the centripetal force on the car. Remember that 1 mph = 0.447 m/s.

Centripetal acceleration =  m/s2
Centripetal force =  N

Explanation / Answer

Given that

A satellite is in orbit about Earth. Its orbital radius is (r) =3.76×107 m.

The mass of the satellite is (ms) =4589 kg

The mass of Earth is(me) = 5.974×1024 kg.

The speed of the satellite is given by v =Sqrt(Gme/(r+h) =Sqrt(6.67*10-11*5.974×1024 kg./(6.38*106+3.76×107 m)=3.009*103m/s

Given that 1 mi/s = 1609 m/s then 1m/s =(1/1609)mi/s

Then the orbital speed of the satellite is v =1.870mi/s

2)

Given that

1 mph = 0.447 m/s.

then the automobile travellling with a speed is given by (v) =56 mph =56(0.447)m/s=25.032m/s

Mass of the automobile(m) =632kg

Radius is (r) =40.1m

The centripetal acceleration is given by a =v2/r

Centripetal force is (F) =mv2/r

Now substitute all the values in the above formula you get the required answer

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