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somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitational pull of the moon. the mass of the moon is 1/81 that of the earth. how far from the center of the earth is this point?

Respuesta :

It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. sylent_reality · 8 years ago
xero099

The point in which both gravitational forces are cancelled is located at 352890 kilometers from the center of the Earth.

We must determine the point between the Earth and the Moon where a particle is at equilibrium, that is, where gravitational forces of the Earth and the Moon have the same magnitude but opposite direction.

By the Newton's Law of Gravitation and the Newton's 2nd and 3rd Laws we  have the following expression:

[tex]\Sigma F = -\frac{G\cdot M\cdot m_{P}}{x^{2}} + \frac{G\cdot m_{P}\cdot m}{(L-x)^{2}} = 0[/tex]

Where:

  • [tex]G[/tex] - Gravitation constant, in newton-square meters per square kilogram
  • [tex]m[/tex] - Mass of the Moon, in kilograms.
  • [tex]m_{P}[/tex] - Mass of the particle, in kilograms.
  • [tex]M[/tex] - Mass of the Earth, in kilograms.
  • [tex]L[/tex] - Distance between the centers of the Moon and the Earth, in meters.
  • [tex]x[/tex] - Distance between the particle and the Earth, in meters.

Now we proceed to simplify this expression:

[tex]\frac{M}{x^{2}} = \frac{m}{(L-x)^{2}}[/tex]

[tex]\sqrt{M}\cdot (L-x) = \sqrt{m}\cdot x[/tex]

[tex]\sqrt{M}\cdot L = (\sqrt{m}+\sqrt{M})\cdot x[/tex]

[tex]x = \frac{\sqrt{M}\cdot L}{\sqrt{m}+\sqrt{M}}[/tex]

[tex]x = \frac{L}{\sqrt{\frac{m}{M}}+1}[/tex]

If we know that [tex]L = 3.921\times 10^{8}\,m[/tex] and [tex]\frac{m}{M} = \frac{1}{81}[/tex], then the distance between the particle and the Earth is:

[tex]x = \frac{3.921\times 10^{8}\,m}{\sqrt{\frac{1}{81} }+1}[/tex]

[tex]x = 3.529\times 10^{8}\,m[/tex]

The point in which both gravitational forces are cancelled is located at 352890 kilometers from the center of the Earth.

To learn more on Newton's Laws, we kindly invite to check this verified question: https://brainly.com/question/13678295

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