1) A spy satellite is in circular orbit around the earth and makes four revoluti

Question

1) A spy satellite is in circular orbit around the earth and makes four revolutions

every day.

a. How high above the earth’ssurface is the satellite?

b. Calculate the satellite’s speed?

and

2) A hand gun fires a 12.0 g bullet at a speed of 400 m/s.

a. What is its kinetic energy?

b. At what speed must a motorcycle of mass 173 kg move to have the same kinetic energy as the bullet? Express your answer in km/h.

3) .A dinner plate of mass 510 g is pushed 60 cm along a dining table by a

constant force of 3.0 N directed 22 ° below the horizontal. If the coefficient of

kinetic friction between the plate and the table’s surface is 0.44, determine

the work done on the plate by

a. the applied force

b. the force of gravity

c. the normal force

Explanation / Answer

Kepler's Third Law: T ² =kR ³. (The square of the orbital period of an object is proportional to the cube of the orbital radius.) The value of k depends only on the mass of the body that the object is orbiting, and is equal to 4Ï€ ²/GM where M is the mass of the body. For Earth, k = 9.9×10^-14 m ³/s ².

(a)

4 revolution a day = 24/4 = 6 hours per 1 revolution
T = 6 h = 21,600 s.
R ³ = T ²/k = (21,600 s) ² / (9.9×10^-14 m ³/s ²) = 4.71×10^21 m ³,
so R = 1.68×10^7 m.
Subtract Earth's mean radius of 6.37×10^6 m to get the altitude of 1.04×10^7 m or 1.04×10^4 km.

(b)

For circular acceleration, a = v ²/r.
v = (length of orbit) / (orbital period)
= 2πR / T = 2π(1.68×10^7 m) / (21,780 s)
= 4850 m/s.
So, a = (4850 m/s) ² / (1.68×10^7 m) = 1.40 m/s ².

(2)

(a)

Kinetic energy = 1/2 Mv^2

= 0.5*0.012*400^2

=960J

(B)

Kinetic energy is same i.e.. = 960J

960=1/2 Mv^2

M = 173 kg

v^2 = 960*2/173 = 11.02

v = 3.3m/s

v = 3.3/1000*3600/1 = 11.99km/h

(3)

a.
As the block is pushed along a FRICTIONLESS
HORIZONTAL table,

applied force=3 X cos22=2.78 N
Therefore, workdone by the applied force= 2.78 X0.60
= 1.66 joule

b. since no vertical displacement is occured,
work done by the normal force,
= the total normal force X vertical displacement
=(0.51 X 9.81 + 3 X sin22) X 0
=0 Joule.

c.

since no vertical displacement was occured,
work done by the gravity force = 0 joule.
d.
work done by net force = work done by horizontal force +
work done by total vertical force
=1.66 + 0
=1.66 Joule

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