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I will openly admit! I am god awful at math! I even used Google to look up how to do this. I checked everywhere I could for legit over an hour for an answer. I cannot even begin HOW to start with this problem and it's due on Wednesday so I'm getting a little stressed out. Here's the problem:

At 11:00pm on September 22nd, 2013 determine the local sidereal time in Abbotsford. Calculate and describe how you can obtain the local hour angle for the star Capella (R.A. 5h 16m 41.5s). Note: Using Stellatium or a similar software program to look up the answer is not acceptable, however, you may use the software to verify your answer.

Help would be greatly appreciated ##### Share on other sites

People around here have a successful day when their shoes end up on the right feet.

FYI.

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People around here have a successful day when their shoes end up on the right feet.

FYI.

You put shoes on your feet... ##### Share on other sites

You put shoes on your feet... I typically wear crocs. Laces involve too much frustration.

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Today I plugged into the USB jack on the first try and it was not by luck either.

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My physics problem that bothers me is as follows:

Ignorig effects from propulsion, if a spaceship takes off from earth going speed of light, and the light reflects off earth the speed of light, if you look behind you would you see anything but earth ever? (You and the light reflecting of earth going same speed.)

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My physics problem that bothers me is as follows:

Ignorig effects from propulsion, if a spaceship takes off from earth going speed of light, and the light reflects off earth the speed of light, if you look behind you would you see anything but earth ever? (You and the light reflecting of earth going same speed.)

wouldn't matter because travelling at the speed of light time does not exist. And as soon as you left you have already arrived at your destination.

So yes until you have arrived at your destination it would be as if you never left.

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My physics problem that bothers me is as follows:

Ignorig effects from propulsion, if a spaceship takes off from earth going speed of light, and the light reflects off earth the speed of light, if you look behind you would you see anything but earth ever? (You and the light reflecting of earth going same speed.)

Doesn't matter, because anything that has mass, i.e.. you & your spaceship, can not move at the speed of light.

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I will openly admit! I am god awful at math! I even used Google to look up how to do this. I checked everywhere I could for legit over an hour for an answer. I cannot even begin HOW to start with this problem and it's due on Wednesday so I'm getting a little stressed out. Here's the problem:

At 11:00pm on September 22nd, 2013 determine the local sidereal time in Abbotsford. Calculate and describe how you can obtain the local hour angle for the star Capella (R.A. 5h 16m 41.5s). Note: Using Stellatium or a similar software program to look up the answer is not acceptable, however, you may use the software to verify your answer.

Help would be greatly appreciated Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

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Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

Wut

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Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

I was totally gonna say that but then I fell down cuzz my shoes were on the wrong feet.

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Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

That awkward feeling when you have no idea if Aladeen is yanking the OP's chain or that is actually a legitimate answer.

Oh well, back to killing my brain cells with beer.

mmmmmmmmmmmm beer.

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Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

Don't any off you have clocks or watches? What is the witchery above?

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Local Sidereal Time:

Sept 22, 2013, 23:00 PST

Translates to

Sept 23, 2013 07:00 Universal Time (UT)

Translates to

a Julian Day of (2456559.79176)

The Julian Day at Noon Greenwich on Jan 1, 2000 = (2451545.0)

2456559.79176 - 2451545.0 = 5014.79176 (number of days (Universal Time) between those 2 days)

Greenwich Sidereal Time (In Degrees) is found as follows:

GST = (Sidereal Time on Jan 1st 2000 in degrees) + (#of Sidereal Degrees/day) (Number of Days)

GST = (280.461)+(360.98564737)(5014.79176) = 1810548.3109093416

Next from that number we want to eliminate multiples of 360

Which leaves us with:

108.31090934

Next we subtract the Longitude of Abbotsford (122.3280) from that number

108.31090934 - 122.3280 = -14.0170966

-14.0170966 + 360 = 345.98290934 degrees (Ascension Angle)

Now we translate that into time,

Hours = (24/360) X 345.98290934 = 23.0655272 Hours

So 23 Hours with a remainder of .0655272

Minutes = Hour Remainder x 60 = 3.931632

So 3 Minutes with a remainder of .931632

Seconds = Minute Remainder x 60 = 55.89

So 55 or 56 seconds depending on if you want to round up

23:03:55

I don't even know how you did that. I'm actually scared to put it down on my paper because if it's right I almost feel like my teacher would be like, "Wow. I didn't even know how to do that and I'm a Physics Master. Can you solve the equation for the meaning of life for me?"

.......And I'll have no idea what I just said. I'm gonna use it anyways. This stupid problem is worth literally 0.1% of my final grade. We have like 4-5 assignments throughout the year, each with around 9 questions, and all 5 assignments combined is worth 10% of our final grade. This 1 question is ridiculous.

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I don't even know how you did that. I'm actually scared to put it down on my paper because if it's right I almost feel like my teacher would be like, "Wow. I didn't even know how to do that and I'm a Physics Master. Can you solve the equation for the meaning of life for me?"

.......And I'll have no idea what I just said. I'm gonna use it anyways. This stupid problem is worth literally 0.1% of my final grade. We have like 4-5 assignments throughout the year, each with around 9 questions, and all 5 assignments combined is worth 10% of our final grade. This 1 question is ridiculous.

i will use my insane brain power to help translate his nonsense into laymen terms. his post actually means now forgive the rough translation (my astrophysics is a little rusty.. But here we go. "a fool once uttered, zero is a constant but with a belt it is an eight." obviously I only translated the relevant parts but the rest was kinda amateurish and I didnt want to embarrass anyone... So feel free to use my translation it is more likely that your professor will understand that a little better.
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Today I learned that Aladeen is smart.

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My physics problem that bothers me is as follows:

Ignorig effects from propulsion, if a spaceship takes off from earth going speed of light, and the light reflects off earth the speed of light, if you look behind you would you see anything but earth ever? (You and the light reflecting of earth going same speed.)

Time slows down relative to how fast to the speed of light you are going, once you go the speed of light (which would require all the energy in the universe) time would stop for you completely relative to the rest of the universe. But practically speaking, there would be no earth to look back at since all the energy of the universe (including the earth and the light reflecting off of it) was put into you going the speed of light, so you would literally be going nowhere since there would be real point of reference.

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I forgot to include the second part of your question this is really easy:

Local Hour Angle (LHA) for Capella:

LHACapella = LSTAbbotsford - RACapella (If the result is negative add 360, if the result is greater than 360 subtract 360)

We can just use our LST degree calculation: 345.98290934 degrees

Next we need to find what the (R.A. 5h 16m 41.5s) translates into degrees

so we just do the last step of our last calculation backwards:

[(41.5 X 360/24) /60] /60 = 0.17291666

(16 X 360/24) / 60 = 4

5 X 360/24= 75

Add all those results together and

Aur RACapella = 79.017291666 degrees

So,

LHACapella = 345.98290934 - 79.017291666 = 266.96517674 degrees