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wild_turkey_willie
Lv 5
wild_turkey_willie asked in Science & MathematicsAstronomy & Space · 1 decade ago

I want to find the point between earth and the sun where the pull of their gravities are equal but opposite?

Since the gravitational force exerted by a body is proportional to its mass and inversely proportional to the square of the distance from the body, at this point

mass of sun/(dist. fr. sun)^2 = mass of earth/(dist. fr. earth)^2, or mass of sun/mass of earth = (dist. fr. sun/dist. fr. earth)^2.

My references all say that mass of sun/mass of earth = 333000, so dist. from sun/dist. from earth = sqroot(333000) = 577.

Then the equilibrium point is 1/578 of the distance between the earth and the sun = 93000000 / 578 = 161000 miles from earth. (Is this okay so far?)

But the moon is always at least 225000 miles from earth, so when it orbits between the earth and the sun, moving in a direction perpendicular to the gravitational pull of both, it is beyond the equilibrium point and the sun is pulling on it harder than the earth is. At these times, why wouldn't the sun tear the moon out of earth orbit? My argument must contain an error - please show me where I have gone wrong. Thanks!

Update:

Thank you for the good answers and information about the Lagrangian points.

Is it then correct that a satellite cannot be placed in a stable orbit around the earth that takes it beyond L1? Is the distance from earth to L1 the theoretical limit for the radius of a stable earth orbit?

Update 2:

(For these questions assume the moon is missing, for it would surely affect a satellite orbiting the earth at such a great distance.)

3 Answers

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  • dogwood_lock
    Lv 5
    1 decade ago
    Favorite Answer

    The Moon doesn't need Earth's gravity to keep it from falling into the sun. The fact that it is orbiting the sun (along with Earth) is enough. The point you are talking about is called the L1 Lagrange Point.

    Your calculations assumed that the objects are not moving relative to each other. Think about this for a second: imagine that the Earth magically disappeared and where the Earth-Moon pair had been we just had the Moon itself. What would happen? The Moon would continue in orbit around the sun, making the trip once every 365 days, just like always. Our solar system would still have 8 planets, but the 3rd planet would be called Luna.

    Yes it's true that when the Moon is between Earth and the sun the gravity of the sun is stronger on it than the gravity of the moon, but both the Earth and the Moon have a significant amount of momentum carrying them in orbit around the sun. You have to allow for this to calculate correctly the location of the Lagrange Point.

    P.S. I just thought of another way to look at this problem. The same logic will show you that, when Earth is between the Moon and the Sun, the sun's gravitational pull on Earth is stronger than the Moon's gravitational pull on Earth, so why doesn't Earth fall into the sun? It's because Earth doesn't need the Moon's gravity to prevent that disaster. Earth's momentum is enough. Ditto for the Moon.

    Imagine that there is a space station at the Earth-Sun L1 Lagrange point. If you put on a space suit and stepped out the airlock on the sun side of the station, you would gradually start drifting toward the sun. But what if a satellite in orbit around Earth came whizzing by right over your head, would it also drift toward the sun? Not necessarily, because it has momentum relative to the station and you don't. So, yes, it is possible for a satellite to have an orbit which takes it inside the L1 point.

    Source(s): http://en.wikipedia.org/wiki/Lagrange_point
  • Peter T
    Lv 6
    1 decade ago

    The Lagrangian points are not the points where the gravitational fields of each body are equal but rather the points where the combined gravitational fields of the two bodies produce the centripetal force that will keep a third body rotating at the same angular speed (and thus be stationary with respect to the two bodies in a reference frame that rotates with them.)

  • jor_rox
    1 decade ago

    What you are describing is call the Lagrangian Point. It's a nifty math model used for satellite positioning...

    here's a nice wiki article:

    http://en.wikipedia.org/wiki/Lagrangian_point

    the math comes out saying that L1 (the point between Earth and the Sun) is about 0.01 the distance to the sun, or

    0.01 * 93,000,000 = 930,000

    So that it why we still have our moon

    Source(s): http://en.wikipedia.org/wiki/Lagrangian_point
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