by Karel Celustka, February 1999
Here's the relevant quantities:
Sorry! There are no new quantities here.
Here are the relevant formulae that we have so far:
a_{c} = V^{2}/r = 4(p)^{2}r/T^{2}
(T_{1}/T_{2})^{2} = (r_{1}/r_{2})^{3}
F_{g} = Gm_{1}m_{2}/r^{2}
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What is the force of gravity between the two 4.2 Kg bowling balls whose
centers are 1.2 m
distant?
Here is what you start with:
What is the force of gravity between the earth and the moon? What acceleration
does the moon
undergo? (Hint: use F=ma) What is its period of motion? (Use a centripetal
acceleration equation)
First, you need to realize again, that^{ }F_{g}
= Gm_{1}m_{2}/r^{2}
and that m_{1 }=29.11*10^24 kg (earth's mass)
and m_{2}
=7.2*10^22 kg (moon's approx. mass). Next,
we need to figure out what the distance is
from the center of the earth
to the center of the moon. Looking this up, we find the crust-to-crust
distance is 3.85*10^8 m,
the radius of the moon is 1.74*10^6 m, and the radius of the earth is
6.37*10^6 m. Now we add
all this up and get that r=3.93*10^8 m. Now we can just put in the
numbers: F_{g} =(6.67*10^{-11}
N*m^{2}/kg^{2})(29.11*10^24
kg)(7.2*10^22 kg)((3.93*10^8
m)^-2).
The answer you should get is 2.03*10^20 N.
For the next part, take the answer from the first part 2.03*10^20
N and divide it by the moon's
mass of 7.2*10^22 kg. You should get [2.03*10^20 N=(7.2*10^22 kg)a]
that a=2.8*10^-3
m/s/s.
Next, we want to use the equation a_{c}
= 4(pi)^{2}r/T^{2}, and we know a=2.8*10^-3
m/s/s. And we
know that r=1.74*10^6 m.
Now, solve this equation. 2.8*10*-3 m/s/s=4(pi)^{2}(1.74*10^6
m)/(T^2). You
should get that T=2.3*10^6 s
or 27 days.
^{Go back to: Table of Contents}
What is the force of gravity on a 10 kg object twice earth's radius above the earth?
First, find the earth's radius. It is 4.7*10^6
m, and since the 10 kg object is twice earth's radius
above the earth, r=1.41*10^7
m. We also need the earth's mass, it is 29.11*10^24 kg. Now,
we just stuff all this into the force of gravity equation. F_{g}
=(6.67*10^{-11} N*m^{2}/kg^{2})(29.11*10^24
kg)(10 kg)((4.22*10^7
m)^-2). You should get that the force of gravity is 10.9 N.
^{Go back to: Table of Contents}
What distance from the center of the moon is the attraction between
a 500 kg object and the
moon itself equal
to 2.5 N?
We know that F_{g} =2.5
N, the m=500 kg, the moon's m=7.2*10^22 kg. Now, stuff all knowns
into the force of gravity
equation and solve: 2.5 N=(6.67*10^{-11}
N*m^{2}/kg^{2})(7.2*10^22
kg)(500
kg)((r)^-2). You
then should find that r=3.1*10^7 m.
^{Go back to: Table of Contents}
^{The answers to each problem follow it in parentheses. They also link to a solution to the problem.}
^{Try the problem, check your answer, and go to the solution if you do notunderstand.}
^{What is the centripetal acceleration of a car going 21 m/s around a corner with a radius of 45 m? (9.8 m/s/s )}
^{What must be the radius of curvature for a curve in a road if cars going 27 m/s never exceed a lateral acceleration of 4.2 m/s/s? (173.6 m)}
^{My econo-box pulls .72 g's (multiply .72 by 9.8) of lateral acceleration going around a corner witha radius of 35 m. What is its speed? (15.7 m/s)}
^{A centrifuge has a tangential velocity of 34 m/s and a radius of 12 cm. What is the centripetal acceleration at its edge? (9633 m/s/s)}
^{A certain car can pull .92 g's of lateral acceleration. What radius turn can it do at 40 m/s? (177.5 m)}
^{Your 1963 microbus can do about 4.5 m/s/s of centripetal acceleration. How fast dare you go around a corner with a radius of 60 m? (8.8 x 107m)}
^{What centripetal force is needed to make a 5 kg hammer swing 3.4 m/s in an arc with radius 1.75 m? (33 N)}
^{What centripetal force is needed to make a 65 kg rider go 15 m/s on the edge of a ride with radius of 4.5 m? (3250 N)}
^{A 320 kg space probe has jets which can exert a centripetal force of 120 N. What is the sharpest radius of a turn it can make if it is going 520 m/s? (7.2*10^5 m/s)}
^{How fast can your 800 kg car go around a corner with a radius of 13 m when the available centripetal force is 6500 N? (10.3 m/s)}
^{What is the force of gravity between the two 4.2 kg bowling balls whose centers are 1.2 m distant? (8.17*10^-10 N)}
^{What is the force of gravity between the earth and the moon? (2.03*10^20 N)}
^{What is the force of gravity on a 10 kg object twice earth's radius above the earth? (10.9 N) Go back to: Table of Contents}
^{The centripetal acceleration=v^2/r. So, (21^2)/45=9.8 m/s/s. (Go To Problem )}
^{We know v=27 m/s, r=? and }a_{c}=^{4.2 m/s/s. Set up the equation and solve: 4.2=(27^2)/r. So r=173.6 m. (Goto Problem)}
^{We know v=?, r=35 m and }a_{c}=^{(.72*9.8) m/s/s. Set up the equation and solve: 7.056=(v^2)/35. So, v=15.7 m/s. (Goto Problem)}
^{We know v=34 m/s, r=.12 m and }a_{c}=^{? . Set up the equation and solve: }a_{c}^{=(34^2)/.12. So, }a_{c}^{=9633 m/s. (Goto Problem)}
^{We know v=40 m/s, r=? and }a_{c}=^{(.92*9.8) m/s/s. Set up the equation and solve: 9.016=(40^2)/r. So, r=177.5 m. (Goto Problem)}
^{We know v=? , r=60 m and }a_{c}=^{4.5 m/s/s. Set up the equation and solve: 4.5=(v^2)/60. So, v=16.4 m/s. (Goto Problem)}
^{We know v=3.4 m/s , r=1.75 m, m=5 kg and F}_{c}=^{? . Set up the equation and solve: F}_{c}^{=(5)(3.4^2)/1.75. So, F}_{c}^{=33.03 N. (Go To Problem )}
^{We know v=15 m/s , r=4.5 m, m=65 kg and F}_{c}=^{? . Set up the equation and solve: F}_{c}^{=(65)(15^2)/4.5. So, F}_{c}^{=3250 N. (Goto Problem)}
^{We know v=520 m/s , r=? , m=320 kg and F}_{c}^{=120 N . Set up the equation and solve: 120=(320)(520^2)/r. So, r=7.2*10^5 m. (Goto Problem)}
^{We know v=? , r=13 m , m=800 kg and F}_{c}^{=6500 N . Set up the equation and solve: 6500=(800)(v^2)/13. So, v=10.3 m/s. (Goto Problem)}
^{Go and see Example #1 (Goto Problem)}
^{Go and see Example #2 (Goto Problem)}
^{Go and see Example #3 (Goto Problem) Go to: Problems Formulas Table of Contents Go back to Tutorial Page}