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Homework HeLP

  1. Mar 5, 2004 #1
    I was just wondering. How come the gravitational field strength of the sun is much larger than the gravitational field strength of the moon at when you calculate both values at Earth's position?
  2. jcsd
  3. Mar 5, 2004 #2

    Doc Al

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    Staff: Mentor

    The sun is much bigger. (About 30 million times as massive as the moon.)
  4. Mar 5, 2004 #3
    Well, it should really be said that the sun is more massive than the moon by a larger amount than the square of the ratio of the distance between the Earth and the Sun and the distance between the Earth and the Moon.

  5. Mar 5, 2004 #4
    Yeah, I was gonna put something like that. What about the force of "attraction" (distance) or something?
  6. Mar 5, 2004 #5
    Newton's Law of Gravitation:

    [tex]F_g = G\frac{m_1m_2}{r^2}[/tex]
    where G is the graviational constant, [itex]m_1[/itex] is the mass of the first body, [itex]m_2[/itex] is the mass of the second, and r is the distance between them.

    The ratio of the two forces are

    [tex]\frac{F_\textrm{Sun on Earth}}{F_\textrm{Moon on Earth}} = \frac{G}{G}\frac{m_\textrm{Sun}}{m_\textrm{Moon}}\frac{m_\textrm{Earth}}{m_\textrm{Earth}}\frac{r_\textrm{Moon to Earth}^2}{r_\textrm{Sun to Earth}^2} = \frac{m_\textrm{Sun}}{m_\textrm{Moon}}\frac{r_\textrm{Moon to Earth}^2}{r_\textrm{Sun to Earth}^2} [/tex]

    You want the ratio on the left-hand-side to be greater than one, so this inequality follows naturally

    [tex]m_\textrm{Moon}r_\textrm{Sun to Earth}^2 < m_\textrm{Sun}r_\textrm{Moon to Earth}^2[/tex]

    which can be rewritten in the form

    [tex]m_\textrm{Sun} > m_\textrm{Moon}\Big(\frac{r_\textrm{Sun to Earth}}{r_\textrm{Moon to Earth}}\Big)^2[/tex]

    which is the ratio I described.


    Edit: Thought this inequality was closest to what I said.
    Last edited: Mar 5, 2004
  7. Mar 5, 2004 #6
    How do I do these 2 questions:

    1. Calculate the net gravitational field strength due to the moon and Earth halfway between Earth (Mass of earth=5.98 times 10 to the 24 kg) and the moon (Mmoon=7.349 times 10 to the 22 kg). The moon's mean orbital radius is 3.845 times 10 to the 8 meters.)

    2. Find the gravitational field strength on the surface of Jupiter. Jupiter's diameter is 1.428 times ten to the 5 km and its mass if 317.83 times that of Earth.
  8. Mar 5, 2004 #7
    Use Newton's Law of Gravitation for both.

    [tex]G = 6.67~\times~10^{-11}\frac{\textrm{N}\cdot\textrm{m}^2}{\textrm{kg}^2}[/tex]

  9. Mar 5, 2004 #8
    I know that....but my English kinda sucks. I wouldn't know which radius or which mass to use....
  10. Mar 5, 2004 #9
    They want the gravitational forces due to the Earth and due to the Moon half-way between the two. The Moon orbits the Earth at, on average, [itex]3.845~\times~10^8\textrm{m}[/itex], so half-way between the two would be half of this number.

    They want the gravitational force due to Jupiter if you were on the "surface" of Jupiter (a slippery term, indeed!). They noted that the diameter (twice the radius) of Jupiter is [itex]1.428~\times~10^5\textrm{km}[/itex]. Divide this in half to get the radius. Don't forget to convert kilometers to meters!

  11. Mar 17, 2004 #10
    Thank you cookiemonster for the help. I've solved the question. Sorry for getting back to you so late. By the way, I need help again....here's the question:

    If you take your electric razor, which has a resistance of 440 Ohms to Europe, where the voltage is twice as high, what factor will the current through the razor change? Will this be of any concern to you? Explain.

    BTW, I just had a quiz on Ohm's Law and Power....I thought I failed it, but I got 100%! *ScArY*
  12. Mar 18, 2004 #11
    The subscript A denotes (in America) and E (in Europe).

    [tex]I_{A} = \frac{V_{A}}{R}[/tex]

    [tex]I_{E} = \frac{2\times{V_{A}}}{R} = 2\times{\frac{V_{A}}{R}} [/tex]

    Now substitute [tex]I_{A}[/tex] for [tex]\frac{V_{A}}{R}[/tex]

    [tex]I_{E} = 2\times{I_{A}}[/tex]
    Last edited: Mar 18, 2004
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