An interplanetary spaceship passes through the point in space where the gravitational forces fro

An interplanetary spaceship passes through the point in...

Key Concepts

Newton's Law of Universal Gravitation

This law governs how any two masses attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is fundamental in calculating the gravitational pull from bodies like the Earth and the sun and is used to determine where—in space—their forces can balance each other.

Gravitational Equilibrium

Gravitational equilibrium occurs when the net gravitational force acting on an object is zero because the forces from other bodies cancel each other exactly. In problems involving two large masses, finding the equilibrium point involves setting the gravitational forces (derived from Newton's law) equal, enabling the calculation of the specific location where these forces balance.

Lagrange Points

Lagrange points are positions in a system of two large orbiting bodies, such as the Earth and the sun, where a smaller object can maintain a constant position relative to the two bodies. The point where the gravitational pulls cancel is one example (specifically the L1 point in the Earth-sun system), and although the net external force is zero there, it does not imply that the position is stable for long-term hovering without active control.

Unstable Equilibrium

An unstable equilibrium is a condition where a slight displacement of an object leads to forces that further move it away from its initial position. Even if an object is placed at a point where the net gravitational force is zero, small perturbations can cause it to drift away, necessitating continuous adjustments or thrust corrections to maintain its position.

Transcript

00:01 So in this question, we're told that an interplanetary spaceship passes through the point in space where gravitational forces from the sun and the earth on the ship exactly cancel out.

00:15 And we're asked to calculate how far from the center of the earth this point is using the data in appendix f.

00:24 And then we're asked to comment on whether or not the spaceship could stay at that point indefinitely.

00:32 If it switches its engines off once it gets to that point in space.

00:37 So let's start with point a.

00:40 So we need to find the point where the gravitational force from the sun on the spaceship is equal to the gravitational force from the earth on the spaceship.

00:55 So let's just go ahead and write in our equation for the gravitational force.

01:00 So we're going to have capital g, the mass of the space.

01:03 Sun times the mass of the spaceship.

01:06 We'll call that just little m divided by the distance between the spaceship and the sun, which actually i'm going to write that in a second.

01:20 And then for the force from the earth, we're going to use capital g, m, e for the mass of the earth, times m.

01:32 And what we're trying to find is r, the separation between the earth and the spaceship.

01:40 So if the distance between the sun and the earth is big r, then the distance between, let's say our spaceship is here, the distance between the sun and the spaceship is going to be big r minus little r.

02:02 So i'm going to write that in there.

02:06 Next we can do some cancellations.

02:08 So we're going to cancel out the mass of the spaceship and capital g.

02:14 And let's do some rearranging.

02:18 So what i'm going to do is i'm going to put all the masses on one side and all of the distances on the other side.

02:37 Now, instead of writing this as r squared over capital r minus r squared, i'm going to just put this all into one big square.

02:52 Where.

02:54 So it's going to be r over r minus r.

02:58 It's all squared.

03:00 And so what that allows us to do is it allows us to create this equation where we have a ratio of the distances on the one side and the ratio of the masses on the other side under this square root.

03:17 And now we can plug in the masses and capital r and solve for little r, which is what we're trying to find.

03:25 So big r, just to remind you, it's the separation between the earth and the sun.

03:31 And we can get that from appendix f, which tells us that that distance is 1 .5 times 10 to the 11 meters.

03:46 And then let's plug in the mass of the earth, so 5 .94 times 10 to the 24 kilograms.

03:57 And the mass of the sun, 1 .99 times 10 to the 30 kilograms.

04:05 And so we can take that whole ratio and take the square root, and we find that we get a value of 1 .732 times 10 to the minus 3 over here.

04:21 Sorry, there's no units there...

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