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Kepler's Law

  • Thread starter Trizz
  • Start date
  • #1
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Homework Statement


(a) Determine the forces that the moon and the sun exert on a mass, m, of water on Earth. Your answer will be in terms of m with units of N. (Use data from table 8-1 in your textbook for this question. The mass of the moon is 7.3 x 10 ^22 kg, and it can be assumed to be 3.9 x 10^5 km from the Earth's center.)

______ m N (force exerted by moon)

______ m N (force exerted by sun)


Homework Equations



Fgr = G M1m2 / r^2

mass of earth = 5.98 x 10^24



The Attempt at a Solution



I thought that I had everything for the equation, but I still can't seem to get it. Check out my work and see if there's something I'm missing. I'm not sure on what to use for the radius.

Fgr = (6.67e-11)(7.3 x 10^22)(5.98 x 10^24) / (3.9 x 10^5)^2
 

Answers and Replies

  • #2
343
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Say the earth is not there but replaced by a bit of water of mass "m". Then?
 
  • #3
41
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Thanks for the response Bright Wang.

So I replaced my equation with m, and this is what I got

Fgr = (6.67e-11)(7.3 x 10^22)(m) / (3.9 x 10^5)^2

then

Fgr = (4.860e12)(m) / (3.9 x 10^5)^2

then

(3.9 x 10^5)^2 = 4.869e12m

does that look right?
 
  • #4
343
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The given data for radius is in KM.
 
  • #5
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Alright I multiplied the radius side by 1000 to make up for the KM units. But one more question. Is that number supposed to be divided by 2 since I want radius, and isn't that diameter? Anyways, i got .03 for an answer, does that seem too small?
 
  • #6
ideasrule
Homework Helper
2,266
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Which number? There's no diameters in this question.
 
  • #7
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i need r^2 for the Force of gravity equation. The statement said that the moon is 3.9 x 10^5 km from the earth. Is that not diameter? or is that the radius? Regardless, i still got .03 for the answer, which doesnt seem to make sense to me
 
  • #8
41
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Actually, im not sure what i got was .03. But here's my equation anyway.

(3.9 x 10^5)^2 = 1.521e11 x 1000 (for km) = 1.521e14

(6.67e-11)(7.3 x 10 ^22)(m) = 4.8691e12(m)

so.....

4.8691e12(m) / 1.521e14 = m????
 
  • #9
343
1
Distance is radius. It doesn't even matter, r represents the distance between the objects.

"
4.8691e12(m) / 1.521e14 = m???? "

What? Your just using the equation.

You want to find Fg. Fg = GMm/r^2.... Why do you have mass on both sides? You trying to find force not acceleration.

Its just plug and chug. I didn't get 0.3m N.
 
  • #10
41
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I thought you said to use "m" for a variable ???



"Say the earth is not there but replaced by a bit of water of mass "m". Then? "

or do i just ignore that number
 
  • #11
343
1
m can be whatever.

Fg = GMm/r^2 = G*(Mass of the moon)*m / distance{in meters}^2 = {some number}*m
 
  • #12
41
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wasnt able to get it in time. Thanks for the help though. Hopefully my teacher will explain it better
 

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