Calculate Gravitational Force at Distance R/2.41 from Star 1

AI Thread Summary
The discussion revolves around calculating the distance from star 1 where the gravitational force on a starship is zero, given that star 2 has twice the mass of star 1 and they are R apart. The initial calculation suggested a distance of 2/3R, but the correct answer is R/2.41, derived from setting the gravitational forces from both stars equal. A participant pointed out errors in the original formula and clarified that the gravitational force from star 1 is GmM/x^2 and from star 2 is Gm(2M)/(R-x)^2. The final equation leads to the conclusion that the starship must be positioned directly between the stars for the net gravitational force to be zero.
grouchy
Messages
72
Reaction score
0
A starship of mass m is traveling between a pair of binary stars that are a distance R apart. the mass of star 2 is exactly twice the mass of star 1. How far from star 1 will the gravitational force on the starship be equal to zero? Express the answer in terms of R.

When I try, I get 2/3R but the answer should be R/2.41

GmM/R^2(x) - Gm2M/R^2(R-x) = 0
GmM/R^2(x) - Gm2M/R^2(R) + Gm2M/R^2(x) = 0
GmM/R^2(x) + Gm2M/R^2(x) = Gm2M/R^2(R)
x = Gm2M/R^2(R)/[3GmM/R^2]

x = 2/3R
 
Physics news on Phys.org
grouchy said:
A starship of mass m is traveling between a pair of binary stars that are a distance R apart. the mass of star 2 is exactly twice the mass of star 1. How far from star 1 will the gravitational force on the starship be equal to zero? Express the answer in terms of R.

When I try, I get 2/3R but the answer should be R/2.41

GmM/R^2(x) - Gm2M/R^2(R-x) = 0
GmM/R^2(x) - Gm2M/R^2(R) + Gm2M/R^2(x) = 0
GmM/R^2(x) + Gm2M/R^2(x) = Gm2M/R^2(R)
x = Gm2M/R^2(R)/[3GmM/R^2]

x = 2/3R
Your whole formula is wrong! You have "R2" in the denominator of both fractions, but have multiplied by x and R-x. I have no idea why you would do that. I assume that "x" is the distance from the starship to the star 1 (it would be good practice to specifically say that) so the gravitational force from the star 1 would be GmM/x^2 and the gravitational force from star 2 would be Gm(2M)/(R-x)^2. Assuming the starship is between the two stars, the will be no (net)gravitational force on the starship when GmM/x^2= 2GmM/(R-x)^2. I would recommend you divide both sides of the equation by GmM immediately, then multiply boths sides by x^2 and (R-x)^2.
 
I don't know why but I keep getting stuck at (R-x)^2= 2x^2
 
You are almost there your final equation that you posted is right-- the rest is math:

(R-x)^2 = 2x^2 \Rightarrow \\<br /> (R-x) = \sqrt{2} x \Rightarrow \\<br /> R = x + \sqrt{2} x \Rightarrow \\<br /> x(1+\sqrt{2}) = R \Rightarrow \\<br /> x = R/(1+\sqrt{2}) \Rightarrow \\<br /> x \approx R/2.41<br />
 
And technically, the force of gravity on something is NEVER 0. Although the Net Force can make it seem so. And it could just get to a point so low that its almost irrelvant. But the gravitational force between objects can't be zero, no matter how far apart they are. So I think you should have mentioned Resultant Force somewhere in there. And are we supposed to assume that the ship is right in between the stars? Because it could be arranged like this :
-> . · . <-
where the left and right dots are stars and the middle dot is the ship. Since the ship is not EXACTLY in between the stars (in this case the ship is higher on the "y-axis"), the distance would probably be different than if the ship was right in between.
 
KalvinDeathX said:
And are we supposed to assume that the ship is right in between the stars? Because it could be arranged like this :
-> . · . <-
where the left and right dots are stars and the middle dot is the ship. Since the ship is not EXACTLY in between the stars (in this case the ship is higher on the "y-axis"), the distance would probably be different than if the ship was right in between.
but then there would be a force composant pointing downwards, hence the net force in such arrangement can't be zero and it follows that the ship must be right in between the stars.
 
Very true, if gravity is the only relevant force.
 
Back
Top