Part A Calculate the period of a satellite orbiting the Moon, 97 km above the Mo

Question

Part A

Calculate the period of a satellite orbiting the Moon, 97 km above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 km.

Express your answer using three significant figures and include the appropriate units.

Part B

A 0.70-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface.

If the cord will break when the tension in it exceeds 45 N , what is the maximum speed the ball can have?

Part C

A car of mass M = 1100 kg traveling at 40.0 km/hourenters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires as shown in (Figure 1) . Use g = 9.80 m/s2 throughout this problem.

What is the radius r of the turn if = 20.0 (assuming the car continues in uniform circular motion around the turn)?

Explanation / Answer

part A
given data
The radius of the Moon is 1740 km.
For "orbit", centripetal acceleration = gravitational acceleration, or
²r = 4²r/T² = GM/r²
T = (4²r³/GM)
where G = Newton's gravitational constant = 6.674*10^11 N·m²/kg²
and M = mass of central body = 7.35*10^22 kg
and r = 1740km + 97km = 1837 km
Then T=sqrt((4*3.14*3.14*1837000^3)/(6.674*10^11*7.35*10^22 ))
T = 7059 s = 1.96 hrs

PartB)
given data
M=0.70 kg
r=1.5m
Ft = 45 N
Ft (tension force) = Fc (centripetal force)
Ft = mv^2 / r
v^2 = Ftr / m
v = sqrt (Ftr / m)
v = sqrt(45*1.5/0.70) = 9.81 m/s

PartC)
given data
M= 1100 Kg
v = 40 km/hr = 40 *5/18 m/s = 11.11 m/s
= 20.0
By equations.
Gravity: Ag = g*sin(Theta)
Centripetal: Ac = (v^2/r)*cos(theta)
g*sin(Theta) = (v^2/r)*cos(theta)
so r = v^2 cost / (g*sin(theta))
r = v^2/[g*tan(theta)]
r = (11.11^2)/(9.80*tan(20))=34.6 m

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