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Meteorite 80,000 km from the Earth is moving towards the earth at 2,000 velocity on impact? (3 points): m/s: Ignoring air friction, what will be its 0.29 x 10'...

Question

Meteorite 80,000 km from the Earth is moving towards the earth at 2,000 velocity on impact? (3 points): m/s: Ignoring air friction, what will be its 0.29 x 10'm/s 2.09* 10 mls 21x105 m/s 1.9x 10' m/s 1.09 x 10'm/sBriefly support your answer (4 pts):

meteorite 80,000 km from the Earth is moving towards the earth at 2,000 velocity on impact? (3 points): m/s: Ignoring air friction, what will be its 0.29 x 10'm/s 2.09* 10 mls 21x105 m/s 1.9x 10' m/s 1.09 x 10'm/s Briefly support your answer (4 pts):



Answers

(II) A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of $5 \times 10^{9} \mathrm{kg}$ . It is approaching the Earth on a head-on course with a velocity of 660 $\mathrm{m} / \mathrm{s}$ relative to the Earth and is now $5.0 \times 10^{6} \mathrm{km}$ away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

So this problem We're looking at application of these radical equations. The equation we're working with his V equals K divided by the square root of d que represents. Uh, K is gonna be a constant in this case is gonna be 3 50. It's given to us in this approach. In this problem, D is the distance that you will prove that the meteorite will travel and V is the velocity of the media, right, as it's approaching Earth. So in this case, we know that K is 300 fifties because it's given to us. We also want we want to find the velocity when the meteorite has traveled 6001 oclock when the meteorite is 6000 kilometers away from the center of the earth. Well, in this case, we already have the equation set equal to V. So all I have to do is solve. I know that 350 divided by the square of 6000 will come out to be approximately 4.5 okay and that is measured in kilometres per seconds.

For this problem on the topic of gravitation, were told that a planet has a mass of seven times 10 to the 24 kg and a radius of 1600 km. It then gravitationally attracts a meteorite that is initially infinitely far away and address relative to the planet. The meteorite falls toward the planet. And we want to find the speed at which it reaches the planet's surface. We can use energy conservation for the situation. So the initial kinetic energy of the planet and meteorite K one plus the initial potential energy you won Is equal to the final Kinetic Energy K two plus the final gravitational potential energy Youtube. And so this is half um V one squared minus the gravitational potential energy initially G. M mm over are one is equal to the final kinetic energy A half M. The two squared minus the final gravitational potential energy. GM capital M. Over our two capital M. Is the mass of the planet and little M. The mass of the meteorite So big m is seven Times 10 to the 24 Kg are two is equal to the radius of the planet, 1.6 times 10 to the six m. And our one is equal to infinity, which means that you one is equal to zero. Now, if we assume that the media starts addressed, V one is also equal to zero. It's okay one plus you one is equal to zero. So the kinetic energy that the meteorite gains is equal to the initial gravitational potential energy between the planet and the meteorite. And so we can rearrange this equation and solve for B to the speed of the meteorite when it reaches the planet's surface. And so we get v. 2 to be too G times the mass of the planet M over the radius of the planet Capital are substituting our values in. We get the speed of the meteorite To be 2.4 Times 10 to the four meters per second.

Yes, conserving the kinetic energy we have half am the final squared minus G mass off art mass of the material divided by readies of the art. So he finds on the surface of the art which equals have em We initiate square the spirit of approach minus g e m All of later the distance at which it is which is our initial. So something this we find fi final equals one point 11.3 kilometers per second. Now, when it is 5000 kilometers above the surface, rip this this Toby R E does 5000 kilometer on. Let's say this is VF prime solving this, you will find VF prime. It was 8.94 kilometers per second.

So in this question, we want to obtain the speed of the meteorite when it reaches the planet's surface. We're told it starts at rest at a distance very far away. So enough to be taken as infinite. We know the formula for kinetic energy is half MV squared and potential energy is minus the gravitational constant, multiplied by the two masses over the separation. So what we know is that the fact that the energy must be conserved from the initial state to the final state, such as the initial kinetic energy, plus the initial potential energy must be equal to the final kinetic energy plus the final potential energy. So we're told also that the initial kinetic energy is equal to zero and the distance at the beginning is taken to be infinity. So as this tends to infinity. Here, you is going to tend to zero. So from this we can tell that K one plus key. U. One is equal to zero. So now we know that K two plus you too must also be equal to zero. Now we can use these expressions here to sub in for K two and Youtube. So this is that half mv t squared minus G. Um Over artie is equal to zero. Okay, so now what we can do is equate these two so a half M V squared equals G. Um um over our two. So what we can do now is cancel out these M terms. And we know that we've been given the velocity, the velocity is what has been trying to find in this question. We've been given the radius is 1600 kilometers. So 1600 times 10 to the three m. We know the masses seven times 10 to the 24. So the gravitational constant is just a constant, has a value of 6.67 times 10 to the minus 11. So we're able to use these to work out a value of V. Two. So V two is equal to the square root of two Gm. Uh huh. Over are, too, which is equal to when these values are subbed into here 2.4 times 10 to the four m per second as our final answer, but when it reaches the surface of the planet.


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