Week 3, Post 2

Worksheet 3, Problem 1

The question asks us to find the difference between the sidereal and solar day on Mars. As the planet orbits the Sun, the right ascension that it was pointing at the day before (towards the center of its orbit, the Sun) would no longer be pointing towards the center. We can see this illustrated in the following drawing:

To solve this, we need to determine the period ($P_M$) of Mars's revolution around the Sun. This will allow us to determine how much of an angle ($\theta$) is covered during Mars's 24-hour rotation period. Then, we can determine how much time it takes to make up for that angle $\theta$ such that the right ascension is once again pointed towards the Sun.

We can calculate $P_M$ using the following relation:

$$\frac{P_\oplus^2}{P_M^2} = \frac{r_\oplus^3}{r_M^3}$$

Where:
$P_\oplus$ is the period of Earth's orbit around the Sun, or a year.
$r_\oplus$ is the distance between the Earth and the Sun.
$P_M$ is the period of Mars's orbit around the Sun, or a year.
$r_M$ is the distance between Mars and the Sun.

Let's solve for $P_M$:

$P_M = (\frac{P_\oplus^2 \cdot r_M^3}{r_\oplus^3})^{1/2}$
$= (\frac{(1 \ year)^2(1.5 \ AU)^3}{1 \ AU}^3)^(1/2)$
$=(1 \ year^2 \cdot 3.375)^{1/2}$
$=(3.375)^{1/2} \ year \cdot \frac{365 \ days}{1 \ year}$
$\approx 670.54 \ days$

Next, we can use the period $P_M$ (the Martian year) to determine the angle $\theta$ that is traveled in a 24-hour period. Note that a 24-hour rotation period is 1 day.

$$\frac{\theta}{360^{\circ}} = \frac{1 \ day}{P_M \ days}$$
$\theta = \frac{360^{\circ} \cdot 1 \ day}{P_M \ days}$

This angle $\theta$ can then be used to calculate the amount of time $t$ it takes to rotate $\theta$ degrees, given that we know Mars has a 24-hour rotation period.

$$\frac{\theta}{360^{\circ}}=\frac{t}{24 \ hours \cdot 60 \frac{mins}{hours} \cdot 60 \frac{secs}{mins}}$$

Let's plug in $\theta$ and solve it.

$t = \frac{\theta \cdot 86400 \ secs}{360^{\circ}}$
$= \frac{\frac{360^{\circ} \cdot 1 \ day}{P_M \ days} \cdot 86400 \ secs}{360^{\circ}}$
$= \frac{\frac{360^{\circ} \cdot 1 \ day}{670.54 \ days} \cdot 86400 \ secs}{360^{\circ}}$
$= 128.85 \ secs$

That's about 2 minutes, which seems reasonable.

Acknowledgment: worked on this problem in class with Awnit and Elton.

Worksheet 3, Problem 2

This question asks us to determine the Local Sidereal Time (LST) at different times. We didn't use any diagrams to solve this question. We did make sure to read the question.

It's important to understand the difference between the solar day and the sidereal day. I found this website to be helpful:

• A solar day is the time it takes for Earth to rotate around its axis. 
• This is measured from noon to noon, so that the Sun appears in the same place in the sky, defined as 24 hours.
• A sidereal day is the time it takes for Earth to rotate around its axis relative to the celestial sphere.
• This is measured by it time it takes a particular star to pass directly overhead (celestial meridian) on two successive nights.

The website included this neat diagram that illustrates the concept (similar to the one we used in Problem 1):

Sidereal day, from http://astronomy.swin.edu.au/cosmos/S/Sidereal+Day

On Earth, a sidereal day takes about 23 hours and 56 minutes, which is 4 minutes less than the solar day. This means that the sidereal day is 4 minutes faster than the solar day, so a given star will rise 4 minutes earlier every night.

Also, as discussed in class, the equinoxes and solstices serve as benchmarks for LST. At noon:

• Vernal Equinox: 00:00:00
• Summer Solstice: 06:00:00
• September Equinox: 12:00:00
• Winter Solstice: 18:00:00

To determine LST at a certain date and time, we use a 3-step process:

1. Find the LST at noon on that date, using the LSTs of the equinoxes and solstices. With each month, LST moves by about 2 hours.
2. Add or subtract the number of hours off from noon of the time that you're concerned with.
3. Add or subtract the number of minutes to account for the difference of 4 minutes between solar day and sidereal day. This means a gain of 1 sidereal minute every 6 hours.

a) What is the LST at midnight on the Vernal Equinox?

At midnight after the Vernal Equinox, we are 12 hours past noon. The LST at noon is 00:00:00, so adding 12 hours brings us to 12:00:00. Add 2 minutes (1 minute/6 hours, so 2 minutes/12 hours) to account for the sidereal day being 4 minutes shorter than the solar day. This gives 12:02:00 LST.


b) What is the LST 24 hours later (after midnight in part 'a')?

Starting from 12:02:00 LST at midnight after the Vernal Equinox. 24 hours (solar day) corresponds to 4 sidereal minutes. Therefore, 4 minutes after 12:02:00 LST (from part 'a') is 12:06:00 LST.


c) What is the LST right now (to the nearest hour)?

Today is February 20, 2019, at 8:00 PM, so 20:00 EST.

Winter Solstice: 18:00:00
Vernal Equinox: 00:00:00

There are 3 months between the Winter Solstice and the Vernal Equinox, so that means there are 2 sidereal hours/month. The Vernal Equinox is on March 20, so we're 1 month away, or 2 sidereal hours. Then, we'll subtract 2 hours from 00:00:00, so we get LST of 22:00:00. 8 PM is 8 hours after noon, so add 8 to 22:00:00 to get 06:00:00. 8 hours is 1/3 of a day, so $1/3 \times 4$ min = 1 min 20 sec. Add this to get 06:01:20 LST.


d) What will the LST be tonight at midnight (to the nearest hour)?

At midnight on that day, we'd start with 22:00:00 LST at noon. Next, add 12 hours to get 10:00:00. Then, add 2 minutes to account for the 12 hours after noon, giving 22:02:00 LST.


e) What LST will it be at Sunset on your birthday?

My birthday is September 9. Let's count it as 6 PM EST (for a dinner party?).

Summer Solstice (June 21): 06:00:00
September Equinox (September 23): 12:00:00

September 9 is 2 weeks (half a month) before September 23. Since there are 2 sidereal hours/month, there is 1 sidereal hour for half a month. LST would be 11:30:00. 6 PM is 6 hours later, so add 6 hours to get 17:30:00. Then, 6 hours is a quarter of a day, which means we gain 1 sidereal minute. This leaves us with 17:31:00 LST at dinner time.


Acknowledgment: worked on this problem in class with Awnit and Elton. Special mention to Elton for edifying us on a systematic approach to this question.

Worksheet 3, Problem 3

This question is tricky, so it's important to read the question.

Recall the definition of local sidereal time (LST): the right ascension that is on your meridian right now.

By definition, we can only see AY Sixteenus at LST of 18:00:00, because its RA is 18 hours. It follows that on any day (including the first of each month) the LST of AY Sixteenus would be 18:00:00.

Therefore, the star would never be visible on the meridian at midnight LST.

General Post 1

The problem of space junk is a growing one; the number of satellites being launched is increasing very quickly - projections predict that there will be over 5,600 commercial launches under 500 kg from 2017 to 2027, up 10 times from the previous decade. The problem is that satellites are very difficult to fix while in orbit, so usually they just have to be brought back down to Earth. However, many old satellites are left in orbit. There are an estimated 750,000 pieces of space junk orbiting the Earth. They travel extremely fast, at 8 kilometers per second, which means a collision could accidentally pulverize an operational satellite. This can cause a cascading chain of collisions known as the Kessler effect, which would leave swaths of low-earth orbit unusable because of the debris. This poses a risk to the future of satellite applications.

A Japanese company called Astroscale proposes a new technology to reduce the amount of space junk in low-earth orbit. Astroscale claims to be the first in-orbit debris capture and removal, which leverages a two stage process: rendezvous and magnetic capture. For their test run, they will launch “chaser” and “target” modules that separate after reaching orbit. The chaser will chase after the target, and then capture it with a magnetic capture mechanism. When the target is docked, the chaser and target will power back toward Earth and burn up on re-entering the Earth’s atmosphere.

https://www.bloomberg.com/news/articles/2019-02-13/as-launches-boom-a-venture-bets-on-bringing-satellites-down

General Post 2

Last week, Opportunity, the NASA Mars Rover, was officially pronounced dead. The rover, a robot the size of a golf cart, was originally planned for a 90-day mission, yet it ended up roving about Mars’s surface for over 14 years. It is especially impactful because it brought about a paradigm shift in space exploration - rather than a stationary lander studying a single spot, a rover can move about and check out interesting things. They also promote a deeper level of engagement with the general public - making it seem as if one were actually “walking on the surface”.

The decision to finally end the mission was not an easy one. Opportunity had been unresponsive since June, because a giant dust storm obscured its solar panels, preventing it from generating enough energy to stay awake. In the fall, NASA was only willing to spend a month trying to reconnect with the rover, but many on the team felt that that was not enough time, leading NASA to allow some more time. However, it became clear that the rover was finally dead, possibly because the solar panels were covered by too much dust to be blown away, or some electronic component failed. We may never know exactly what killed Opportunity, until one day when we have astronauts on Mars to inspect the intrepid rover.

https://www.nytimes.com/2019/02/13/science/mars-opportunity-rover-dead.html