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# Tutor profile: Sheila S.

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Sheila S.
Astrophysics student, researcher
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## Questions

### Subject:Astronomy

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Question:

Say we have a spaceship and we want to get it from Earth to Jupiter as quickly as possible, using as little energy as possible. How long would it take? (The Earth is 1 AU from the Sun, and Jupiter is 5.2 AU from the Sun.)

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Sheila S.

A minimum energy trajectory is called a Hohmann transfer. We are going to treat this trajectory as a planet that has a semimajor axis $$a$$ equal to the average of the distances of Earth and Jupiter from the Sun. So, our "perihelion distance" is 1 AU, and our "aphelion distance" is 5.2 AU. Our semimajor axis is the average distance of our "planet" from the sun, or $$\frac{1 AU + 5.2 AU}{2} = 3.1 AU$$. Now, we can use Kepler's third law $$p^2 = a^3$$ to find the orbital period $$p$$ of the planet. When we plug in $$a$$ into this equation, we get $$p$$ = 5.5 years. To reach Jupiter, we just need half of one "orbit", because we're only going from Earth to Jupiter and not back again. So $$\frac{5.5 yrs}{2} = 2.75$$ years.

### Subject:ACT

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Question:

ACT MATH: What is the value of x when $$4x + 6 = 6x – 8$$ ?

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Sheila S.

First, we notice that all the coefficients (the numbers before the $$x$$s) AND all the the lone numbers are even. This means that we can divide the whole equation by two; this makes it simpler to solve. To do this, we just divide all the numbers we see by two, and we get $$2x + 3 = 3x - 4$$. Now we want to try to get all the $$x$$s on one side and all the other numbers on the other side. We're going to choose that the numbers go on the left side and the $$x$$s go on the other side, but it works both ways. So first, we add 4 to each side, and we get $$2x + 7 = 3x$$. Now, we want to subtract $$2x$$ from each side, and we get $$7 = 5x$$. Now, we divide both sides by 5, because we want a statement that says $$x =$$ some number. So, we divide by 5 and get $$x = \frac{7}{5}$$.

### Subject:Physics

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Question:

SPECIAL RELATIVITY: Let's say we have a spaceship leaving earth, and we are on earth. (We denote t to be OUR time and x the distance of the spaceship from the Earth in OUR frame.) The spaceship’s equations of motion are $$x(\tau) = \frac{c^2}{a} (cosh(a\tau/c) − 1)$$ and $$ct(\tau ) = \frac{c^2}{a}(sinh(a\tau/c))$$, where $$\tau$$ is the time in the spaceship's frame. The spaceship accelerates at a rate of $$a = 10 m/s$$ for 1 earth year ($$t=1$$ year). We know the speed of light is $$c = 3x10^8 m/s$$ and the number of seconds in a year is approximately $$3x10^7 s$$. How far from us will the spaceship be when it stops its acceleration?

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Sheila S.

We know that after 1 year of acceleration (t = 1 year = $$3x10^8 s$$), the spaceship will have traveled $$\tau$$ time in its own reference frame. We can take our equation $$ct(\tau ) = \frac{c^2}{a}(sinh(a\tau/c))$$ and divide both sides by $$c$$, so we get $$t(\tau ) = \frac{c}{a}(sinh(a\tau/c))$$. Now, since we know that $$t(\tau)$$ = $$3x10^7 s$$, and we know that $$a = 10m/s$$, we can plug in those values and solve for $$\tau$$. We plug in to get $$3x10^8 = \frac{c}{a}sinh(\frac{a\tau}{c})$$, and solving for $$\tau$$, this becomes $$\tau = 2.6x10^7 s$$. Now we know $$\tau$$ in the ship's frame of reference, and we can plug it into our equation for $$x(\tau)$$, which is the distance traveled in the stationary frame: we have $$x(\tau) = \frac{c^2}{a} (cosh(a(2.6x10^7/c) − 1)$$, and solving for $$x$$, this becomes $$3.87x10^15m$$. This is 0.41 light years.

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