Circular Motion, Orbits, and Gravity Problems Include all equations used Include

Question

Circular Motion, Orbits, and Gravity ProblemsInclude all equations used

Include Free Body Diagrams, as well as any equations used and how you calculated out the answers:

1. A typical modern wind turbine has blades 56 m long that spin at 13 rpm. At the tip of the blade, what are (a) the speed and (b) the centripetal acceleration?

2. In recent years, astronomers have planets orbiting nearby stars that are quite different from planets in our solar system. Kepler-39b, has a diameter that is 1.2 times that of Jupiter, but a mass that is 18 times that of Jupiter. What would be the period of a satellite in a low orbit around this large, dense planet?

Earth mass: 6.0x1024 kg Earth radius: 6.4x106 m

Jupiter mass: 1.9x1027 kg Jupiter radius: 7.0x107 m

3. A 75 kg man weighs himself at the north pole and at the equator. Which scale reading is higher? By how much? Assume the earth is a perfect sphere. Explain why the readings differ.

Thank you so much!

Explanation / Answer

1) v = w r = (13*2*pi/60)*56= 76.2 m/s

b) a = v^2/r = 76.2^2/56= 103.7 m/s^2

2)

G M m/r^2 = m v^2/r

v^2 = G M/r

v = 2 pi r/T

(2* pi *r/T)^2 = G M /r

4 pi^2 r^2/T^2 = G M/r

T^2 = 4 pi^2 r^3/( G M)

T = 2*pi*sqrt((1.2*6.4E6)^3/(6.67E-11*18*1.9E27))= 88.5 s

3) at the north pole

N - mg = 0
so N = 75*9.81= 736 N

but at equater

N - m g = mv^2/r

N = m g + mv^2/r

so difference = mv^2/r = 75*(2*pi*6.4E6/(24*60*60))^2/6.4E6= 2.54 N
dm = 2.54/9.81 = 0.26 kg

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.