Physics, 19.07.2019 02:50 davidaagurto

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?

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

EarthPMoon

"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

Answer from: Quest

t = 1/400 h ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.

Answer from: Quest

$p_{i}$ = initial pressure

$p_{f}$ = final pressure = 2 $p_{i}$

$v_{i}$ = initial volume

$v_{f}$ = final volume = $p_{i}$ /4

$t_{i}$ = initial temperature = 20 c = 20 + 273 = 293 k

$t_{f}$ = final temperature = ?

using the equation

$\frac{p_{i}v_{i}}{t_{i}}= \frac{p_{f}v_{f}}{t_{f}}$

$\frac{p_{i}v_{i}}{293} = \frac{(2p_{i}) (v_{i}/4)}{t_{f}}$

$\frac{1}{293} = \frac{0.5}{t_{f}}$

$t_{f}=10$

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Somewhere between the earth and the moon is a point where the gravitational attraction of the earth...