Mathematical SOS!

I'm not good at maths. Don't get me wrong, it's cool and all but just I suck at it. Now, take this ability at maths and combine it with the fact that I've been given two (what I see as hard) mathematical questions to answer and one has a little problem. So I was kinda hoping that some kind soul at EOFF could put a poor person out of their number induced misery and help me out. I'm really at the end of my tether with these questions, I don't even know where to begin in trying to tackle them. If anyone can flex their big sexy mathematical muscles and help me out, I would be very grateful!

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Originally Posted by Question 1

1. When it is time for a Space Shuttle to come in for a landing, it needs to re-enter the Earth's atmosphere at the right time and place so that it can land successfully at the designated landing strip. Depending on the Shuttle's altitude, it's de-orbit burn must occur for the correct length of time for the Shuttle to begin its descent at the right speed and in the correct location. The de-orbit burn is done against the direction of travel. The Shuttle keeps going in the same direction but slows down due to the drag on the spacecraft as it enters the atmosphere.

De-orbit manoeuvres are usually done to lower the perigee of the orbit to 60 miles (or less). The Orbiter is captured and re-enters as it passes into the atmosphere at this altitude. There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth.

Determine the change in velocity (delta-V) that the Shuttle will need to make if it is at an altitude of 220 miles above the Earth at apogee and 215 miles above the Earth at perigee, and needs to drop the perigee to an altitude of 60 miles.

Your answer needs to be in feet per second.

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Originally Posted by Question 2

2. Using the change (delta) in velocity that must be used to lower the perigee to a 60-mile altitude (This was your answer to the Shuttle math question for Lesson 1) and assuming the Orbiter's OMS engines have a combined force (thrust) of 12,000 lbs and the Shuttle has a weight of 250,000 lbs (with a full cargo bay), use the equations below to compute the length (or time) of the burn necessary in minutes.

f = ma force equals mass times acceleration and t = v/a time equals velocity divided by acceleration

Your acceleration will be in G's, where 1 G = 32 feet per second per second (this is how far an object travels due to the force of gravity in a vacuum).

Hint: You can also use English slugs instead of G's. 1 slug = 32 pounds. This makes the equation somewhat simpler. Mass of the Orbiter = 250,000/32 slugs.

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... what the hell are you doing freaking astrophysics for if you're not good at maths?

Does math make you happy? It should.

I like math but I'm not a rocket scientist.

I'm not a maths or physics student I'd just like to know the answers to these questions and understand how the answers were obtained. It was a challenge issued by one of my slightly more eccentric teachers, a challenge that I'd like to meet!

Maths that I can do makes me happy.

So these questions are hard then? I thought it was just my inability at Maths!

Ask again in a few years, I'm only a first year on my astrophysics degree and we haven't even done any astro yet :D But yeah, they are hard so it's not you not being able to do them. ;)

Also, S.I. units rock.

LOL Dimensional Analysis ftw. But I don't know any of these formulae, so I can't help.

Those questions appear to require calculus, and I'm definitely out of practice.

They just need to inform me how those mice are doing in zero G. The rest is academic and I'll take their word for it.

Question 1.

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[mike]: When ignorance reigns, life is lost says:
i would think the answers would be (220-60)*2 and (215-60)*2

Ashley says:
Could you just give a quick explanation of how you come to that conclusion?

[mike]: When ignorance reigns, life is lost says:
well, it says that for every 1 mile change there is a 2fps change in velocity

[mike]: When ignorance reigns, life is lost says:
and you need to move from 220 to 60 miles altitude

[mike]: When ignorance reigns, life is lost says:
so that's 220-60

[mike]: When ignorance reigns, life is lost says:
and for every mile you change 2fps

Question 2

[mike]: When ignorance reigns, life is lost says:
a = f / m = 12000 / 250000

[mike]: When ignorance reigns, life is lost says:
then

[mike]: When ignorance reigns, life is lost says:
t = v / a = (v from the prev question) / (a from prev calculation)

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So says the hot astrophysics boy across the street. :love:

I had to tell him I was on a scientific message board to save face though :shifty:

I think my head is going to explode. :jess:

YAAY! Thank you so much Chemical and hot astrophysics boy! :D

It's not too difficult to do the math, mostly physics problems are just hard to read. Chemicals' astrophysics boi looks like he's on the right track anyways.

so cute of them giving you the formulas.

Anyways, take what you got in part one (I'm thinking that if you just do the perigee like it asks it's like 310f/s) and that's v. v/t (t is what you're solving for first, it'll be in seconds) is equal to the force exerted by the thrusters divided by the mass of the spacecraft. I'm not sure why they cared about using Gs, since you want to find time but whatever.

310/t = 12000/250000
12000t/250000 = 310
310*250000/12000 = t

t(seconds)/60 = t(minutes)


Disclaimer: I am a dumb high school student and also I have never worked a physics problem in English units. THere may be some weird rules that don't apply in metric. *shrug* weight and mass and force are not the same :(

[edit-- hurray happy endings. Now who wants to do my problems? ^_^
An .85g bullet is shot at 3.00kg block of wood sitting on a counter 1.20m high. If the bullet is embedded in the block and they land 5.0 m from the counter, what was the initial speed of the bullet?]

Hot astrophysics boy I think made a slight error. He is using the apogee in his calculations, when he should be using the perigee. So it should be (215 - 60) * 2, not 220. And Crystal is right, most physics problems are just hard to read. That and you might not be understanding all of the terms, I had to look up apogee and perigee, haven't heard those in quite some time.

Edit: Also, be sure to convert your acceleration from G's to fpsps before you equate the time, which everyone seems to be forgetting. And to convert your time to minutes.

I can't even understand the question. My calculus is kinda basic.

Edit: Why can't you just use SI units?

i love maths,. but thats just too much writing for me to even bother

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Originally Posted by Chemical

So says the hot astrophysics boy across the street. :love:

I had to tell him I was on a scientific message board to save face though :shifty:

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ROFL!!

That was my favorite method to "study" when I was at Uni. Asking the hot guy next to me :p Too bad he went for Mechanics while I went for Electronics. There were no hotties in my class after that. :mad:

PS: Got married to the hot geek mentioned above though! :love: