<span>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. </span>
sylent_reality
· 8 years ago
Larry Finkelstein, Norman Fischer, and Cassius Schwartz have been overlooked, in my opinion.
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
