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Can someone help me with this please?..

Started by Allana, Jun 19 2005 07:33 PM

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#1 Allana

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Posted 19 June 2005 - 07:33 PM

When it is time for the 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, its deorbit burn must occur for the correct length of time in order for the shuttle to begin its descent at the right speed and in the correct location. The deorbit burn is done against the direction of travel. The shuttle keeps moving in the same direction, but it slows down due to the drag on the spacecraft as it enters the atmosphere. Deorbit maneuvers are usually done to lower the perigee of the orbit to 60 miles (or less). The orbiter will get captured and will re-enter as it passes into the atmosphere of this altitude.

Question :

Calculate the change in velocity (in m/s) that must be used to lower the perigee from 230 miles to a 60 mile altitude? To reduce the shuttle altitude by 1 mile there must be 1 foot per second chane in velocity. (1 f = 0.31m)

Help would be much appreciated

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#2 linds

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Posted 19 June 2005 - 07:34 PM

Ouch... Physics... Brain... Melting... Can't... Focus.

Erk.

Sorry.

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#3 Allana

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Posted 19 June 2005 - 07:35 PM

I feel exactly the same...only another 50 similar questions to go

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#4 linds

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Posted 19 June 2005 - 07:39 PM

Ew, poor you. =/ I did higher Physics, got an A (by appeal, I must admit). It never made much sense to me. I just learned it for the exam without actually understanding.

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#5 Allana

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Posted 19 June 2005 - 07:42 PM

QUOTE(Lindsay @ Jun 19 2005, 07:39 PM)

Ew, poor you. =/ I did higher Physics, got an A (by appeal, I must admit). It never made much sense to me. I just learned it for the exam without actually understanding.

oooo well done...as u got an A, get helpin me with that question!

hehe just kiddin. I really don't understand physics at times. What did u get for maths?

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#6 broughy

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Posted 19 June 2005 - 07:42 PM

don't panic. i'm doing physics at uni & even i dunno where to start with that. i'd guess along the lines of equations of motion, but other than that i'm lost!

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#7 The Wedge Effect

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Posted 19 June 2005 - 07:42 PM

QUOTE(Allana @ Jun 19 2005, 08:33 PM)

When it is time for the 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, its deorbit burn must occur for the correct length of time in order for the shuttle to begin its descent at the right speed and in the correct location. The deorbit burn is done against the direction of travel. The shuttle keeps moving in the same direction, but it slows down due to the drag on the spacecraft as it enters the atmosphere. Deorbit maneuvers are usually done to lower the perigee of the orbit to 60 miles (or less). The orbiter will get captured and will re-enter as it passes into the atmosphere of this altitude.

Question :

Calculate the change in velocity (in m/s) that must be used to lower the perigee from 230 miles to a 60 mile altitude? To reduce the shuttle altitude by 1 mile there must be 1 foot per second chane in velocity. (1 f = 0.31m)

Help would be much appreciated

Um...no ideas. I forgot everything to do with physics. Sorry! Ask Werlop!

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#8 linds

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Posted 19 June 2005 - 07:43 PM

I got an A for that as well, but again it's been a year since I've done it.

I'll have a look at that question. Then ask my brother who's doing sciences at uni, haha!

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#9 Allana

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Posted 19 June 2005 - 07:45 PM

QUOTE(Lindsay @ Jun 19 2005, 07:43 PM)

I got an A for that as well, but again it's been a year since I've done it.

I'll have a look at that question. Then ask my brother who's doing sciences at uni, haha!

whoa thats great. I'll have to work hard this year, i'm not the type who goes into an exam having done very little revision and sails through it....ah well we can dream can't we.
And thanks for your help

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#10 Allana

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Posted 19 June 2005 - 08:17 PM

aaaaanyone?

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#11 Ally

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Posted 19 June 2005 - 08:19 PM

You're best asking in the Physics forum.

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#12 D_n_y

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Posted 19 June 2005 - 08:32 PM

I think the answer to this question is actually quite simple.

i.e To reduce the orbit by 1 mile then you have change speed by 1ft per second.

Therefore 170 miles (230-60) = 170ft per second.

1ft= 0.31m

so 170x0.31= 52.7 m/s

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#13 coca

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Posted 19 June 2005 - 08:34 PM

QUOTE(D_n_y @ Jun 19 2005, 09:32 PM)

I think the answer to this question is actually quite simple.

i.e To reduce the orbit by 1 mile then you have change speed by 1ft per second.

Therefore 170 miles (230-60) = 170ft per second.

1ft= 0.31m

so 170x0.31= 52.7 m/s

That's what I got, too

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#14 Allana

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Posted 19 June 2005 - 08:40 PM

QUOTE(D_n_y @ Jun 19 2005, 08:32 PM)

I think the answer to this question is actually quite simple.

i.e To reduce the orbit by 1 mile then you have change speed by 1ft per second.

Therefore 170 miles (230-60) = 170ft per second.

1ft= 0.31m

so 170x0.31= 52.7 m/s

god, i feel thick now

Thanks a lot though

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#15 D_n_y

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Posted 19 June 2005 - 08:41 PM

no probs

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#16 Dave

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Posted 19 June 2005 - 09:27 PM

thought that isnt for physics is it

Its for NASA no less

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#17 Allana

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Posted 19 June 2005 - 09:29 PM

QUOTE(Dave @ Jun 19 2005, 09:27 PM)

thought that isnt for physics is it

Its for NASA no less

It sure is

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#18 Gavers

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Posted 20 June 2005 - 03:44 PM

Rocket science explained the easy way - one bit goes boom, the other bit goes whoosh the other way. All you need to know

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#19 The Wedge Effect

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Posted 20 June 2005 - 04:15 PM

Yeah. Unfortunately, the physicists makes boom and whoosh complicated, even giving them units.

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#20 Allana

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Posted 20 June 2005 - 04:19 PM

QUOTE(Wedge37 @ Jun 20 2005, 04:15 PM)

Yeah. Unfortunately, the physicists makes boom and whoosh complicated, even giving them units.

Don't mock my inability to use my common sense!

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