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Kepler and Newton

  • Thread starter 3dsmax
  • Start date
  • #1
3dsmax

Homework Statement


a. Planet "P" has a mass that is 2.89 times that of the Earth, an equatorial radius that is 1.89 times that of the earth and is located 7.8 times further from the Sun than the earth is. What is the amount of force applied to the planet by the Sun?

b.what is the orbital velocity of planet "p". What formula do i use for that???

c.What is the period velocity of Planet "P"?

Homework Equations


a. Fg=6.67 e-11 (m1m2)/r^2

c. T^2=r^3

The Attempt at a Solution



NOTE THE EDIT: Fg= 6.67 e-11 [(1.99e30 x (2.89)(5.28e24)] / [(7.8)(1.5 e11)]^2

Fg= 1.48 e21
(but i know the answer is 1.68 e 21)
What did i do wrong


FIXED THIS ONEc. t^2 = 7.8^3
t= 21.8 years
 
Last edited by a moderator:

Answers and Replies

  • #2
hage567
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5.28e24 is not the mass of the earth.
5.98324 first of all is not the radius of the earth, and second is not nearly the right magnitude. But anyway, what you want for this is the mean radius of earth's orbit, since it says the planet is 7.8 times further from the sun than the earth is.
I think you just need to get your numbers straight, then try it again.
 
  • #3
3dsmax
okay well i fixed my numbers but and still getting the wrong answer
 
  • #4
hage567
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Never mind, I see your edit.
 
  • #5
hage567
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According to my sources, the mass of the Earth is 5.98x10^24 kg, not 5.28x10^24 kg. That is where the difference is.
 
  • #6
3dsmax
well i figure it all out the only other problem that i am having now is on this:

1. A 55.0 kg person drives a 2300 kg car and cruises over the top of a hill with an 84.0 meter radius. With what minimum velocity can the car move so that the drive will feel weightless?

I used the formula mu = v^2/r/g
i figure since there has to be no normal to feel weightless so i figured there has to be no friction too. So set it equal to like 0. So then i did like v^2/84m/9.81. Can i do that or is there another way?
 
  • #7
hage567
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I don't understand the "mu". For the driver to feel weightless, the force of gravity must equal the centripetal force. So you can set them equal and solve for v.
 

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