A rogue band of colonists on the moon declares war and prepare...

Question

A rogue band of colonists on the moon declares war and prepares to use a catapult to launch large boulders at the earth. Assume that the boulders are launched from the point on the moon nearest the earth. For this problem you can ignore the rotation of the two bodies and the orbiting of the moon. What is the minimum speed with which a boulder must be launched to reach the earth? Hint: The minimum speed is not the escape speed. You need to analyze a three-body system.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A group of scientists on Mars are planning to send a package to earth using a powerful launcher. Assume that the package is launched from a point on Mars that is the closest to earth for this problem. The rotation of both planets and the orbiting of Mars should be ignored. What is the minimum speed at which a package must be launched? In order to reach earth? Note that the gravitational constant is equal to 6.67 multiplied by 10 to the power of negative 11 cubic meters per kilogram multiplied by second square mass of the earth is equal to 5.98 multiplied by 10 to the power of 24 kg mass of Mars is equal to 6.39 multiplied by 10 to the power of 23 kg distance between earth and Mars is equal to 2.23 multiplied by 10 to the power of eight kilometers radius of Mars is equal to 3.39 multiplied by 10 to the power of 6 m. And the distance reached by the package from the center of Mars is equal to 5.50 multiplied by 10 to the power of 10 m. Awesome. So it appears that our end goal, what we're ultimately trying to solve or what we're trying to figure out is we're trying to determine what the minimum speed at which this package has to be launched. In order for it to reach earth, we're trying to figure out what this minimum speed is in order to launch the package from the middle of Mars to earth. So what is this minimum speed that it has to be in order to achieve this goal of launching the package to earth? So now that we know that we're trying to solve for the minimum speed value, let's read off our multiple choice answers to see what our final answer might be. And let us note that they're all in the same units of kilometers per second. So A is two, B is five C, is 10 and D is 15. OK. So first off, let us write down all of our known variables and under unknown variables. So we know that the gravitational constant, which we're gonna call this value G or specifically capital G. So the gravitational constant is 6.67 and it's multiplied by 10 to the power of negative 11 and once again, its units are cubic meters per kilogram multiplied by seconds squared. We also know the mass of earth which we're gonna call this value. Me and the mass of earth is equal to 5.98 multiplied by 10 with the power of 24. And its units will be kilograms. We also know the mass of Mars which we're gonna call this value. Mm and the mass of Mars is equal to 6.39 multiplied by 10 to the power of 23 and its units will be kilograms. We do not know what the mass of the package is which we're gonna call this value MP. So we do not know what the mass of this particular package is. We also do not know what V is. We do not know what the initial launch velocity is. This value is unknown to us. And this is our final answer that we're ultimately trying to solve for. However, we do know the distance between Earth and Mars, which we're gonna call this value rem and rem which once again, this is the distance between earth and Mars is equal to 2.23 multiplied by 10 to the power of eight kilometers. But we have a bit of a problem all of our other units for radius or for distance values, I should say like the radius is a distance value and the distance achieved are reached by the package and center mars and meters. So as you could see, most of our distance values are in units of meters and same with our gravitational constant has a meter unit in it. So we need to convert kilometers to meters in order to keep all of our units consistent. So I'm gonna go ahead and give you the conversion to save some time, but you can use dimensional analysis to solver the conversion yourself or you could quickly look it up. But rrem value. So rem value in units of meters is 2.23 multiplied by 10 to the power of 11 m. So once again, I've just given you the conversion but you can use dimensional analysis or you just look this up. OK. So moving right along here and we also know one last value, we know the value of R which is the distance reached by the package from the center of Mars and this is equal to 5.50 multiplied by 10 to the power of 10 meters. Awesome. So now our next step is is we need to equate the sum of the initial and final potential and kinetic energies of this particular system which as we should recall, we should be able to write this equation as U I plus K I is equal to UF plus A F which once again, this is saying that the po initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. We also need to note note that sense KF equals zero, we can therefore recall and use the following equation to solve for V which in this case, V is the launch velocity, which is the final answer that we're ultimately trying to solve for. So we will determine that V is equal to the square root of two multiplied by capital G multiplied by me all multiplied by one divided by rem minus capital RM minus one divided by Rem minus R plus two, multiplied by capital G multiplied by mm all multiplied by one divided by capital RM. So one divided by capital RM minus one divided by R fantastic. So now all we need to do is we need to plug in our known values into our equation. So we need to plug in all of our known values into our equation and sol per V which is our final answer. So when we do that, we will find that V is equal to the square root of two multiplied by the gravitational constant, which is 6.67 multiplied by 10 to the power of negative 11. And let us note that the units are cubic meters per kilogram multiplied by second squared. And this is multiplied by our M value which M the mass of earth is equal to 5.98 multiplied by 10 to the power of 24 kg. And we need to multiply this by one divided by Rem, which Rem is 2.23 multiplied by 10 to the power of 11 which this is once again, this is from above answer. So this is one of the variables that is given to us by the pro itself and its units are meters minus capital RM, which capital RM is equal to 3.39 multiplied by 10 that the power of six meters and RM in this case, capital R is the radius of Mars Awesome. Just to quickly recap here. And then we need to subtract this expression which let's add another bracket. So we don't get confused about what's what and we need to subtract this by one divided by which is, are we gonna be now we're on to our R so one divided by re minus R. So we know that Rem is equal to 2.23 multiplied by 10 to the power of 11 meters. And we need to subtract this value by R which we know that R is equal to 5.50. And we need to multiply this by 10 to the power of 10. So awesome. So 10 to the power of 10 and then don't forget that our units are in this case meters. So we're running out of room here, but we're still under the square root symbol. We need to add now two multiplied by capital G which once again, capital G is our gravitational constant, which we need to be able to note that our gravitational constant once again is 6.67 multiplied by 10 to the power of negative 11. And its units are cubic meters divided by or per I should say. So, cubic meters per kilogram multiplied by seconds squared multiplied by the mass of Mars, which we know that the mass of Mars is 6.39 multiplied by 10 to the power of 23 kilograms. Awesome. And then what up above, we forgot to add our bracket to note the end of our equation. So now moving right along. So we need to multiply the mass of Mars by our next expression which is one divided by capital R and minus one divided by lowercase R. So bracket in parenthesis. So one divided by RM, capital RM, which once again, that's the radius of Mars. So that's 3.39 multiplied by 10 to the power of 6 m. Awesome. And we need to subtract this value by one divided by and it's gonna be one divided by R which we know that R is 5.50 multiplied by 10 to the power of 10 m. Awesome. So once we plug in that giant monster of an equation, we will find that the final answer has to be equal to run into the nearest whole number 5010 meters per second. But we need to quickly convert meters per second to kilometers per second because all of our multiple choice answers are in kilometers. So I'm just gonna give you the conversion, but you can use dimensional analysis to find this value. But when you convert this to kilometers per second, and you round to the nearest hole number which will be five kilometers per second. And that's it. That's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B, five kilometers per second. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Gravitational Potential Energy

Kinematics of Projectile Motion

Three-Body Problem

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