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# An alien spaceship traveling at 0.600c toward the Earth launches a landing craft with an advance guard of purchasing agents and environmental educators. The landing craft travels in the same direction with a speed of 0.800c relative to the mother ship. As observed on the Earth, the spaceship is 0.200 1y from the Earth when the landing craft is launched. (a) What speed do the Earth-based observers measure for the approaching landing craft? (b) What is the distance to the Earth at the moment of the landing craft’s launch as observed by the aliens? (c) What travel time is required for the landing craft to reach the Earth as observed by the aliens on the mother ship? (d) If the landing craft has a mass of $4.00 \times 10^{5} \mathrm{kg}$ , what is its kinetic energy as observed in the Earth reference frame?

## A. 0.946$c$B. 0.160 $\mathrm{ly}$C. 0.114 yrsD. $7.50 \times 10^{22} \mathrm{J}$

Gravitation

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##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

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### Video Transcript

it's exercise. We have an alien ship that's moving with a speed V equal to 0.6 times. It's been alive relative to the earth, towards the earth and at a certain time, the ship the sides to lunch, a landing craft to the Earth with that has a speed of 0.8 times the speed of light. See, I'm coming that speed to be prime relative to the ship itself. And at the time when the launching occurs, the distance between the Earth and the ship is too light Years in question. A. We have to calculate what is the speed measure for the landing craft by an observer on the earth. So first, we need to remember the relativistic transformation of velocity form. So you have that you use the speed of the landing craft relative to the earth is equal to the speed of the landing craft relative to the ship, plus a speed of the ship. Nobody noticed that since both are moving in the same direction, I'm adding them, divided by one plus the prime times v develop Assi Square. So this is 0.8 C plus 0.6 c divided by one plus 0.8 times your 0.6 c squared, divided by C square. So this is equal to 0.9 46 times this beautiful light. Then an question be we need to company that this is between, ah, the ship and the earth at the moment of the launching of the landing craft, as observed by an alien that's inch inside this patient. So notice that l the distance in the, uh the earth reference frame is too light ears. Hey, this is the proper distance. That's the distance measure by the still observer on the earth and meaning to coordinate help prime. And since we have the proper distance, we have to divide L. By God, Kang Gamma is the speed off the earth relative to the, uh, the the alien spaceship. And that's you're going six times the speed of light since the speed of the eight and relative to the earth is 0.6 times the speed of light. So this is equal to two like here. So it's two ears times. See, uh, divided by, uh, actually. But if I go and got my is one over the square root of one minus 0.6 square, so this is equal to zero point 16 light years. Then, in questions, see, you have to calculate how much time passes between the launching of the leading craft and it reaching the Earth in the reference frame of the aliens. So first, remember that the distance between the launching site and the earth, according to the aliens, is to your point 16 light years. And also you need to calculate the speed with which the two bodies that is your ending in the landing craft approach each other. So I have here the earth, and here we have the landing craft. Now, according to the According to the spaceship of the agents, the speed of the Earth V is 0.6 times is beautiful act, and the speed of the landing craft is your 0.8 times but alive. So they approach with a speed you of veep Leslie Price. And so the time is just help prime divided by you. So it's syrup or, in 16 light years, divided by 0.6 but a 0.8 oats 1.4 times. It's kind of like so the time is 0.4114 years. Finally, we have information that the landing craft has a mess off four times 10 to the fifth kilograms and I need to find what is the kinetic energy of the landing craft in the reference frame off the earth. Remember that the kinetic energy is able to go on minus one times. EMC squared ever guys. One over the square root of one minus B squared over C squared so K is equal to one over the square root of one minus the square. Don't notice that the speed is there a point nine for six times. Bit of light time C squared Yvette Obey C Square Mine is one times the mass That's four times into the fifth kilograms times he squared just three times 10 to the eighth meters per second squared. So OK is able to 7.5 times 10 22 years which concludes our exorcists

#### Topics

Gravitation

##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp