CIRCULAR MOTION
(questions inspired from "Physics for future presidents",by Richard A. Muller)


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13) REmember Fg = G m M / r2 = m g
m is the mass of an object attracted by a big mass M (m could be your mass and M could be the mass of the EArth)
g is the acceleration due to gravity. On EArth, g1 = 9.8ms/s (sea level)
A) Using G m M / r2 = m g , can you solve for g ? g = ___________  
(use M, G and r of course)

B) g is proportional to _______________ but inversely proportional to ______________________

C) Let's compare the gravity on the Moon g2 to the gravity on the EArth g1.
The Moon has a mass M2= 1/81 the mass of the EArth M1.
( M2 = M1/81 ) (decreasing  M, decreases g)
The Moon has a radius R2 = 3.7  times smaller than that of the EArth R1 (R2 =R1/3.7)
(decreasing r, increases g)
How is the acceleration g2 on the moon compare to the acceleration g1 on EArth?
g2 (moon) = _________ g1 ( earth)
(one way to do it is to find the ratio of g1 / g2  using the formula above and using R1 = 3.7 R2 and M1 = 81 M2 )

D) So How is your weight on the Moon compared to your weight on the EArth ?
Don't worry the attraction is still large enough to bring you back to Moon after a big jump.
That would not be the case if you are standing on a small asteroid, a jump and off you go.

E)  advanced (Math, math, every one loves Math)
Suppose you are standing on an asteroid with radius = 1km.
The density of the asteroid is the same as Earth. (mass = density x volume and volume is proportional to the cube of the radius)
How is the acceleration due to gravity g2 on the asteroid compared to g1 on Earth ?

g1 = ________ g2
So you weight will be divided by ____________, so a small jump and off you go.

14) (cavendish experiments)
A) two  3kg lead balls are placed with their centers 7.25m apart. What gravitational force exists between then ?
Is the force small ?
B) two balls have their centers 2m apart. One has a mass of 8kg. THe other has a mass of 6kg. What is the gravitational force  between them?
C) without computation, if you multiply the distance by 100, the force is ___________ by ________________


















 

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