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QUESTION 45 (3 marks)One asteroid travels in an elliptical orbit and another asteroid travels in cucular orbit about the Sun_ For each asteroid shown below, circle ...

Question

QUESTION 45 (3 marks)One asteroid travels in an elliptical orbit and another asteroid travels in cucular orbit about the Sun_ For each asteroid shown below, circle the alphabet-labelled points where the speed of the asteroid in its orbit is the same_WSuNM

QUESTION 45 (3 marks) One asteroid travels in an elliptical orbit and another asteroid travels in cucular orbit about the Sun_ For each asteroid shown below, circle the alphabet-labelled points where the speed of the asteroid in its orbit is the same_ W Su N M



Answers

An asteroid of mass $m$ is in a circular orbit of radius $r$ around the Sun with a speed $v$. It has an impact with another asteroid of mass $M$ and is kicked into a new circular orbit with a speed of 1.5 $v$.What is the radius of the new orbit in terms of $r$?

6.71. So we have an asteroid moving in a circular orbit around the sun with some speed be, and then it has an impact with a another asteroid of the mass of Capital M, and it's kicked into a new orbit. It has a speed 50% greater than its old one, and we want the fine what its new the new radius of its orbit is. So for an object that's orbiting around something, we know that it's heat is going to be G times the mass of that object, the sun, in this case, over our this square root of all that's and our new V will call, the crime is equal to 1.5. So 1.5 claims she times. The mass of the sun divided by the original radius is equal to the square root of G times, the mass of the sun provided by our new radius. And so solving this for our prime. You get that it's hard over 1.5 squared or zero point for for times are

As we all know that the value of he is equal to under route to G. M. By uh solving it further, I can write the value of M is equal to r. V E squared by two G. And here I can write, the value of art is equal to 365.1 kilometer, which is equal to 365.1 Multiplication tend to depart three m. Now, just putting the value in the application, I can write the value of em is equal to 365.1 Multiplication 10 to the about three. My duplication, 319.2 square by To multiplication 6.67 multiplication tend to the par -11, which on simplification finally, I get the value of mass of the asteroid edge 2.789 multiplication 10 to about 20 Kg.

So we're going to use the conservation of mechanical energy As well as equation 1321. And we can say that the in case of one minus G times the larger mass, smaller mass divided by the separation are are someone equaling case of two minus again, G larger mass, smaller mass divided by our sub two. Now we can for part A we're simply plugging in ah case of one arse of one and a case of too are so too plugging that in. We know the we know the gravitational constant and we know the mass of earth. So we can then solve for M. M. Is then going to be approximately equal To 1.0 Times 10 to the 3rd kilograms for part B, solving for V. This would be equal to the square root of two times the kinetic energy divided by the mass M. And this will be then approximately equal to 1.5 times 10 to the third meters per second. Of course, at our equaling What's given in the problem 1.94, 5 Times 10 to the 7th m. That is the end of the solution. Thank you for watching.

In this problem, we have an asteroid which is 500 km in radius and has an acceleration at the surface equal to 3m/s squared. And we want to determine um first of all, the escape velocity. So for all of these questions, there's three parts. But for all these questions, we're going to be using conservation of energy. So total energy will always equal to the kinetic energy plus the gravitational potential energy. The kinetic energy is going to be equal to have M V. Squared, where V. Is the velocity of the object and the potential energy is equal to negative um G which is the gravitational constant mass of the object, or mass of the two objects, multiplied by each other divided by the distance between them and that's also equal to um MGH or negative MGH where G is the acceleration due to gravity and H is a height above the surface, so or the height above the point where the gravitational excavation is being applied from. And we also know that acceleration from Newton's law of universal gravitation is equal to GM over R squared AM here is the mass of the planet should be a big um figure. Thanks. So for escape velocity, we're going to use a conservation of energy. So our energy on the surface, which is going to be a kinetic energy on the service. So whenever two and B squared, what's the gravitational energy on the surface, which I'm gonna say minus, um th seeing as we know the acceleration on the surface, that's going to be equal to our kinetic energy at the maximal distance. So when it says escape velocity, it means it will never come back. So and it's a minimum velocity needed to never come back. So the kinetic energy to never come back. Okay, semantic energy needed to get the leftover kinetic energy when you have escaped Is going to be equal zero. And the reason for this is because if it's a minimum speed to escape, There should be no speed left over after escaping. So really here is equal to zero and this is added to the gravitational potential energy, which I'm going to say is minus she mm over there. And if we escape and never come back, that means we're always going away and that will make radius the distance away from the planet or the distance away from the center of the planet equal to infinity six years ago to infinity and that exists zero as well as the writer brand Fendi is approximate zero. So when we simplify this, we can get one of the two M V squared would be, here is an escape velocity minus the gravitational energy on the surface, MGH is equal to zero As this is zero and this is zero. So now we have to do is solve this. We can cancel atoms and move everything to the other side. And we can say the is equal to the square root Of two. That's multiplying by. What's that about you? G H where G is the acceleration on the surface and that's three m per second squared two times 3 times our age, which is a distance away from The source of the energy, which is 500 km. So that's 500,000 m. And if you plug this into the calculator, you should get B is equal to one point 73 or 1.7 m per second. So at 1.7 kilometers for a second, it's a big difference. So our velocity is equal to 1.7 km/s. That's our escape velocity. Now, that's the problem is how fast do we have to go? Okay, alright, well, we reach If we start with the velocity of one km/s or 1000 m/s. So in this case the this is the second part of question. Sorry, your B is equal to 1000 meters per second and we're going to do the same thing. So we're going to use our conservation of energy. So are staring energy which is on the ground Is 1/2 MBB squared. Well, B is equal to 1000 m per second. That's what we're launching at minus MGH, which is a gravitational potential energy at the surface. And that's going to be equal to our final energy, which is when, which. And this is how high do we get? So that means all of our kinetic energy has been turned into gravitational potential energy. So I'm going to go ahead and use this equation here which is that is equal to negative G. And I am over our. And the reason we can't use MGH is because we don't actually know the acceleration which will have at this point. The acceleration due to gravity is dependent on distance and the distance away from the planet the asteroid has changed. So we don't actually acceleration here. So we're going to simplify this and see what we can do so we can get rid of and which is a member of the object that we're throwing each case And we can say 1/2 we squared which is 1000. So one half time is 1000 just a second where minus G. H. Which is again three times 8 churches. 500,000 isn't equal to negative G. M over our. And we want to sell for our so all we need to figure out is GM do this will bring down our acceleration equation where A. Is equal the Gm over R squared equal the G. M over I square. Now this we don't know the acceleration or the radius at maximum point. But since Gm is constant, if we calculate the Gm At the starting point is the same as a GM in the ending point. So we use the starting point for this where a is equal to three at the acceleration on the surface Physical the Gm divided by our spread which is 500 1000 squared or you can say Gm which is what we're looking for here is equal to three times 100,000 squared. And if you sell this you get E. M. Is equal to 7.5 Times 10: 11. And if you know the gravitational constant, you can calculate here the mass of the asteroid. But we're not looking for that since we have G. M. G. M. Here. And that's the same as GM here we can substitute 7.5 times 10 to 11. So if this is not a clear 11, check the next thing better. Um and we will consult our so we're gonna go out and do that. So 1/2 Times 1000 Squared minus we Times 500,000 is equal to negative 7.5 times 10 to the 11 divided by our and now we just have the one valuable so you can role play on that side and divide it. I am gonna move ahead because it will take a second. You can do that completely with the calculator now that you have are isolated and what you'd get if you saw our is our is equal to 2.5 times 10 to the five. Um It is one 250 km. Now. For the last part one determine how fast we'll be going if we fall my height of 1000 kilometers. So in this case are starting energy is completely gravitational. So that's negative Gm over. All right. And that will be converted completely into kinetic, which is one over to um b squared. Um Plus whatever energy we have at the surface, whatever gravitational injury is still on the surface, which is three chances. No, this is MGH again. So, mass times. Well, it's got nice here as well. Mass Times G, which is three times are higher acceptance, which is 500,000. And we're solving for me this time. So we can cancel it. I'm like always. And we know GM is 7.5 times 7 11. now, radius were said we're starting at a height above the surface of It's going to the planet starting at height above the surface of 1000 km. However, that's not the height from the gravitational center, which is an extra 500 kilometers. So are here is equal to 1500 cameras. So we're gonna play this in 7.5 times 10-11 divided by 1,500,000 m. It's equal to 1/2. We square. That's what we're solving for minus three times 500,000. And we can we arrange this to solve a beer. The negative 7.5 times 10-11, divided by 1.5. So it kind of tend to go to six Z equals outside plus three times 500 I was in. And all of that will be multiplied by two. and then we take the square root of it and I don't solve A B. Okay. And when we saw the V we should get an answer, velocity is equal to 1,414 m/s. And as you can see, although we started 1000 km away from the surface, well, we're of course going to be coming down with velocity less than our escape velocity. So I hope this was helpful. Thank you.


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