At a certain instant the earth the moon and a stationary...

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×10^5 km in length. Part A Find the magnitude of the net gravitational force exerted on the spacecraft by the earth and moon. Part B Find the direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. Part C What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.

Transcript

00:01 Here i'll be looking at an example where we have to determine the vector nature of the gravitational force using newton's expression for the magnitude, starting with that.

00:14 We also have more than one object in exerting a gravitational force on our object of interest, and so we will have to find a net force by adding up two separate forces.

00:29 And the idea is we have a satellite shown at the bottom left of equilateral triangle, and the earth and the moon are set on the other two vertices of that triangle.

00:43 So we have two forces that we will be calculating with the magnitude given by newton as f is equal to big g, product of two masses.

00:55 I usually write it as big m times a little m, but some books have it as m1 times m2, and then divided by r squared, where r is the separation between the centers of the masses.

01:12 And this works well if your masses are point masses or spheres, which we are assuming here.

01:22 Moon and earth are spheres, the satellite spacecraft, for all intents and purposes is a point.

01:31 So we have two magnitudes that we want to calculate, and i am just going to calculate one of the magnitudes, the force on the spacecraft due to the earth.

01:43 Big g, by the way, is a constant, no matter where you are in the universe, we hope, 6 .67 times 10 to the minus 11, newton's times meter squared per kilogram squared.

02:00 So we're going to take the mass of the earth, multiply by the mass of the spacecraft, and divide by the side of the equal -autical triangle squared.

02:15 So let's put all that in, and we'll use si units for everything.

02:24 Mass of the earth is a well -known quantity.

02:30 The spacecraft mass is given, and we know the distance between the earth and the moon, which is the a.

02:38 That is given.

02:42 And we do want to write that in meters and not kilometers.

02:47 And working that out, that comes out to, let me check it again with big exponents like this.

02:54 It's easy to make a mistake.

03:05 And we still get 3 .13 neutrons.

03:16 Now for the magnitude of the force on the spacecraft due to the moon, i am going to actually take the ratio of the mass of the moon to the mass of the earth.

03:29 And then i can scale the force of gravity from my previous calculation using that figure.

03:40 That is slightly over 1%, 1 .23 times 10 to minus 2.

03:52 And because we're just switching out the mass of the earth with the mass of the moon, we can use that ratio.

04:10 So we can use our calculation that we did before and then multiply it by the scale factor, the mass of the moon times the mass of the earth.

04:23 And that's going to make things a little bit easier with the numbers.

04:28 You can check it by plugging in all those numbers again, but i usually make a mistake when i start working with big exponents, bigger, small exponents, quite the same problem.

04:42 And so we get 3 .85 times 10 to the minus dutons.

04:54 Now that's just the first step.

04:56 Now we have to turn these into vectors.

05:00 And i chose a coordinate system where the largest force, so if i drew these, i would have a really large force pulling that spacecraft to the earth.

05:13 And i don't know if i could draw this in scale, but a very small force, 1 -100th that size, pulling the satellite to the moon.

05:26 And i want to take components.

05:32 I purposefully set up the biggest force along the x -axis, and the smaller force then is at a 60 -degree angle.

05:41 And now i can take components.

05:45 Using magnitude and direction.

05:51 And we need components of both the forces.

05:55 The force of the earth on the satellite.

05:58 The spacecraft has no y component, and it is fully all x.

06:11 The second force on the spacecraft from the moon has two components, the 3 .85 times 10 to the minus 2 times the cosine of 60 degrees, that force is up into the right, the positive x component, and a positive y component.

06:37 So we will use sign and cosine of the angle that that force makes to the x -axis.

06:59 And we're just doing the calculator math.

07:10 And now for the superposition.

07:12 F total, we're just going to add up, these are vectors, so our total force is also going to have an x and a y component.

07:28 That just comes from adding the two columns.

07:31 Do not combine the columns, they are separate components...