Re: help - gravity problem
Joe Pfeiffer
> I thought I understood basic gravity problems but the following high
> school physics problem from my daughter has me stumped ( I think)
> Q. a disabled ( meaning of disable not defined) satellite of mass
> 2400kg is in orbit at a ht of 2000 km above the earth at a speed of
> 6900 m/s. ( my calc show that is exaclty the speed required for a
> circular orbit at that ht). it then says the satelite falls to a ht
> of 800 km calculate what the new speed at the lower ht..
> well I simply calculated the gain in potential energy ( PE = delta
> GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2)
> as the satellite must speed up. and added this to the original speed
> of 6900 m/s to get 10870 m/s , but I am not sure that this correct.
> it certainly doesnt give me the answer in the school text book of 7900
> m/s
>
> ( I used G=6.67E-11, M =5.98E24 kg and r= 6.38E6 m.
> using this the loss in potential energy = 1.9E10 J
>
> I suspect I am going wrong somewhere in not accounting for the fact
> velocity is a vector quantitiy. Surely it must depend on the direction
> the satellite is heading initally. is this really a solvable problem?
There has to be some more information for the problem to be meaningful:
the satellite isn't going to fall to a height of 800 km without
something happening to cause it. You've made one good assumption as to
what the question may have meant (without checking your arithmetic, your
approach looks correct to me), and the "right" approach isn't immediately
clear to me.
(FWIW, I tried just recalculating the orbital velocity at a height of
800 meters and got 7460, which is closer but also clearly not what they
had in mind).
Ken S. Tucker
> I thought I understood basic gravity problems but the following high
> school physics problem from my daughter has me stumped ( I think)
> Q. a disabled ( meaning of disable not defined) satellite of mass
> 2400kg is in orbit at a ht of 2000 km above the earth at a speed of
> 6900 m/s. ( my calc show that is exaclty the speed required for a
> circular orbit at that ht). it then says the satelite falls to a ht
> of 800 km calculate what the new speed at the lower ht..
> well I simply calculated the gain in potential energy ( PE = delta
> GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2)
> as the satellite must speed up. and added this to the original speed
> of 6900 m/s to get 10870 m/s , but I am not sure that this correct.
> it certainly doesnt give me the answer in the school text book of 7900
> m/s
>
> ( I used G=6.67E-11, M =5.98E24 kg and r= 6.38E6 m.
> using this the loss in potential energy = 1.9E10 J
>
> I suspect I am going wrong somewhere in not accounting for the fact
> velocity is a vector quantitiy. Surely it must depend on the direction
> the satellite is heading initally. is this really a solvable problem?
It's ok, some specific info is lacking.
How does a disabled satellite retro fire?
((BS baffles brains))
It sounds like you are expected to find the orbital
velocity at 800km when an impulse was applied to
the sat at 2000 km, to retro fire into an elliptical
orbit with perigee at 800 then retro fire again to
circularize. Anyway assuming you had help from
pink fairies, the standard circular orbit equation is,
Vo = sqrt( GM/r ) .
Ken
Robert Heller
I have a question (knowing practically zip about orbital machanics). If
the satellite was passively in its 2000 km orbit would this orbit
(eventually) decay (friction from cosmic dust or something)? Maybe the
high school program is thinking of this sort of situation. In this case
the 'pink fairies' would just be an entropy effect.
>
>
--
Robert Heller -- 978-544-6933
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Dr J R Stockton
ba.googlegroups.com>, Sun, 7 Jun 2009 21:56:02, dotcom
<tfm...@iprimus.com.au> posted:
>I thought I understood basic gravity problems but the following high
>school physics problem from my daughter has me stumped ( I think)
>Q. a disabled ( meaning of disable not defined) satellite of mass
>2400kg is in orbit at a ht of 2000 km above the earth at a speed of
>6900 m/s. ( my calc show that is exaclty the speed required for a
>circular orbit at that ht). it then says the satelite falls to a ht
>of 800 km calculate what the new speed at the lower ht..
>well I simply calculated the gain in potential energy ( PE = delta
>GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2)
>as the satellite must speed up. and added this to the original speed
>of 6900 m/s to get 10870 m/s , but I am not sure that this correct.
>it certainly doesnt give me the answer in the school text book of 7900
>m/s
>
>( I used G=6.67E-11, M =5.98E24 kg and r= 6.38E6 m.
>using this the loss in potential energy = 1.9E10 J
>
>I suspect I am going wrong somewhere in not accounting for the fact
>velocity is a vector quantitiy. Surely it must depend on the direction
>the satellite is heading initally. is this really a solvable problem?
No, because we don't know why it fell.
Three extreme cases can be modelled as
(1) Gentle Atmospheric drag; its new speed is that for the lower
circular orbit.
(2) It suddenly lost some speed, so that it is in an elliptical orbit
apogee 2000 km perigee 800 km.
(3) It suddenly lost all speed, is coming straight down, and is passing
800 km now.
Also,
(A) It bounced off something elastically, so that its elliptical orbit
has perigee 800 km and speed crossing 200 km is circular speed for that
height.
(B) It bounced off something elastically, is coming straight down, and
is passing 800 km now.
Probably one of the first two is intended; the first for a mid-range
student or the second for an advanced one.
The first thing is to get the exact question as asked. I was once
requested by a young teenager to give the formula for a tree. First
question to ask : is this chemistry/biology, is it engineering, is it
topology (carbohydrate; Euler's Strut, Euler's polyhedrons). She seemed
satisfied to count branches, leaves, and vertices.
== == ==
The moderators' system is not working as it claims that it will. I got
messages; I sent the replies which, the message said, would stop the
system mailing me again; but it did not have that effect. It has an
obvious design fault.
--
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Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
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Ken S. Tucker
Here's a ref on circular orbital speed,
http://ceres.hsc.edu/homepages/classes/astronomy/spring99/Mathematics/sec10.html
and I'll suggest an examination of lunar recession,
http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html
Those are .edu sites and look good to me. The latter shows
that the Moon is receeding from the Earth.
Ok, orbital decay of a 2000 km orbiting sat is nearly
impossible to predict, (IMHO), because there are many
tiny effects (sometimes called perturbations) to be
considered. One would even have to know the shape and
material of the satellite and specific orbit, to calculate
atmospheric drag, magnetic field effects, solar wind,
effect of the moon and so forth.
Regards
Ken
Ken S. Tucker
> On Jun 9, 8:45 am, Dr J R Stockton <reply0...@merlyn.demon.co.uk>
> wrote:
>
> > In sci.space.tech message <be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000v
> > ba.googlegroups.com>, Sun, 7 Jun 2009 21:56:02, dotcom
> > <tfm...@iprimus.com.au> posted:
>
> > >I thought I understood basic gravity problems but the following high
> > >school physics problem from my daughter has me stumped ( I think)
> > >Q. a disabled ( meaning of disable not defined) satellite of mass
> > >2400kg is in orbit at a ht of 2000 km above the earth at a speed o
> f
> > >6900 m/s. ( my calc show that is exaclty the speed required for a
> > >circular orbit at that ht). it then says the satelite falls to a ht
> > >of 800 km calculate what the new speed at the lower ht..
> > >GMm/r) and equated this to the gain in kinetic energy ( 0.5 mv^2)
> > >as the satellite must speed up. and added this to the original speed
> > >of 6900 m/s to get 10870 m/s , but I am not sure that this correct.
> > >it certainly doesnt give me the answer in the school text book of 7900
> > >m/s
>
> > >using this the loss in potential energy 1.9E10 J
> > messages; I sent the replies which, the message said, would stop the
> > system mailing me again; but it did not have that effect. It has an
> > obvious design fault.
>
> > --
> v6.05 MIME.
> > Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms
> & links;
> > Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm
> , etc.
> > No Encoding. Quotes before replies. Snip well. Write clearly. Don't Ma
>
>
> Thanks Dr John
> I quoted the problem verbatim from my daughters text book. you post
> some interesting scenarios, here are my thoughts on them
> 1. re atmospheric drag. My understanding is that atmospheric drag
> would be negligible at 2000km .
> 2. how could it lose some speed?( lets say it had a retro rocket
> firing then in that case you wouldnt be able to solve the
> problem anyway because there is an unknown energy source being
> applied. the way the problem is posed you
> could not possibly get an answer unless you assume only the force
> of gravity is involved.
> 3. If it suddenly lost all speed and was coming straight down then
> again it would involve another force other than gravity.
> so not solvable
> Terry
Well Terry the problem is a bit tough for HS students,
what I roughly figured is the question was aimed at,
http://en.wikipedia.org/wiki/Hohmann_transfer_orbit
I think the textbook author and teacher were being too
ambiguous (unless there is additional context) and that's
either unfair to the student (frustrating), or I dare suggest
incompetence. As a teacher, I wouldn't approve of such a
question, because it's a bit of a "turn-off" in a subject
(orbital mechanics) that needs some confidence to master.
Part of the mission is to encourage students to develope
space technology skills.
Regards
Ken
Steve Willner
dotcom <tfm...@iprimus.com.au> writes:
> ...high school physics problem...
> Q. a disabled ( meaning of disable not defined) satellite of mass
> 2400kg is in orbit at a ht of 2000 km above the earth at a speed of
> 6900 m/s. ( my calc show that is exaclty the speed required for a
> circular orbit at that ht).
I think you got the right answer but were probably misled by the
coincidence in that extra calculation you did. The question is
unclear, but nothing you state suggests the satellite is actually in
a circular orbit.
> it then says the satelite falls to a ht of 800 km
I think here they are suggesting -- but not stating -- that no forces
other than gravity act on the satellite. So the orbit is NOT
circular.
> calculate what the new speed at the lower ht..
> well I simply calculated the gain in potential energy ( PE = delta
> GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2)
> as the satellite must speed up. and added this to the original speed
> of 6900 m/s to get 10870 m/s
This is the right approach, but as you noticed in a later post, you
can't add speeds. Instead add the potential energy change to the
original kinetic energy, and work out what speed corresponds to that
kinetic energy. The mass of the satellite is irrelevant, as you will
probably notice.
As others noted, if non-gravitational forces act, the problem as
stated is incomplete.
--
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
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Dr J R Stockton
vbc.googlegroups.com>, Tue, 9 Jun 2009 06:09:45, dotcom
<tfm...@iprimus.com.au> posted:
>I quoted the problem verbatim from my daughters text book.
In that case, assuming that you actually understand and mean "verbatim",
move the daughter to a different school - one that uses a better grade
of textbook. What you gave cannot be verbatim from a well-written book.
To go further, we need the exact words, warts and all, of the book,
without any intermingled comment from yourself or your daughter. The
title, author, ISBN and page reference could also be useful; some here
may have access to the book and be able to see more context than it
would be proper to copy here.
Your daughter can copy the words, and you can certify it as a true copy.
See signature below.
= = =
I predict that I will again receive a false message from the moderation.
--
(c) John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.
Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Proper <= 4-line sig. separator as above, a line exactly "-- " (SonOfRFC1036)
Do not Mail News to me. Before a reply, quote with ">" or "> " (SonOfRFC1036)
Erik Max Francis
> Dr J R Stockton wrote:
>> In sci.space.tech message <97af933c-beda-4dba-aa2e-81364d4202ef@e20g2000
>> vbc.googlegroups.com>, Tue, 9 Jun 2009 06:09:45, dotcom
>> <tfm...@iprimus.com.au> posted:
>>
>>> I quoted the problem verbatim from my daughters text book.
>>
>> In that case, assuming that you actually understand and mean "verbatim",
>> move the daughter to a different school - one that uses a better grade
>> of textbook. What you gave cannot be verbatim from a well-written book.
>>
>
> circular orbit at 2000km (apparently, I haven't checked), the question
> doesn't say that the satellite is in such an orbit. It does say that it
> falls to a height of 800km.
>
> So the reasonable assumption is that it's in an orbit that allows it to
> be at 2000km at one point in time, and 800km at another.
That would be my guess, too; it sounds asking what the speed of the
satellite will be if it has a perigee of 800 km altitude and an apogee
of 2000 km altitude. It's a bit glib and not terribly clear, though. I
agree with the others that it's not a very useful question, especially
for high school students.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis
Most men do not mature, they simply grow taller.
-- Leo Rosten
David M. Palmer
<be0aecfa-e255-449f...@n4g2000vba.googlegroups.com>,
dotcom <tfm...@iprimus.com.au> wrote:
> I thought I understood basic gravity problems but the following high
> school physics problem from my daughter has me stumped ( I think)
> Q. a disabled ( meaning of disable not defined)
Presumably it doesn't have any propulsion systems so it is in free-fall.
> satellite of mass
> 2400kg is in orbit at a ht of 2000 km above the earth at a speed of
> 6900 m/s. ( my calc show that is exaclty the speed required for a
> circular orbit at that ht).
I haven't done the calculation, but assuming that you did it correctly:
If the orbital speed for a circular orbit is 6900 m/s at 2000 km
height, that does not mean that an object with that height and that
speed is in a circular orbit.
Suppose you climb a 2000 km tall ladder (from Antarctica so you can
ignore Earth's rotation and are initially motionless), and then fire a
rifle that has a muzzle speed of 6900 m/s. Every bullet you fire will
have a 6900 m/s speed at 2000 km height. If you fire it horizontally,
then you will have a circular orbit. If you fire it straight up then
it will have a long, skinny orbit that intersects Earth before it gets
all the way around. If you fire it straight down it will have the same
orbit, only it will complete even less of an orbit. If you fire it at
an angle, then it will have a perigee below 2000 km and a apogee above
that height. It will, of course, travel faster at perigee and slower
at apogee.
If the apogee is below 800 km, then you can complete the question. It
turns out to have the same answer for all orbits with an apogee below
800 km (including the shoot-straight-down case).
> it then says the satelite falls to a ht
> of 800 km calculate what the new speed at the lower ht..
> well I simply calculated the gain in potential energy ( PE delta
> GMm/r) and equated this to the gain in kinetic energy ( 0.5 mv^2)
> as the satellite must speed up. and added this to the original speed
> of 6900 m/s to get 10870 m/s , but I am not sure that this correct.
> it certainly doesnt give me the answer in the school text book of 7900
> m/s
>
> ( I used G6.67E-11, M 5.98E24 kg and r 6.38E6 m.
> using this the loss in potential energy 1.9E10 J
You can't take E 1.9e10 J -> 3970 m/s and add that to 6900 m/s to get
the new speed.
You have to take
Ekinetic_start 5.7e10 J
+ Epotential_change 1.9e10 J
Ekinetic_end 7.6e10 J
And 7.6e10 J -> the 7900 m/s that's the textbook answer.
Basically, you have to add the energies, not the velocities.
Which means that you have to add the square of the velocity and the
square of the KE-equivalent velocity and take the square root of the
sum. (The phrase used is 'add in quadrature')
>
> I suspect I am going wrong somewhere in not accounting for the fact
> velocity is a vector quantitiy. Surely it must depend on the direction
> the satellite is heading initally. is this really a solvable problem?
Yes, the stated problem is a conservation of energy problem, and
solvable.
--
David M. Palmer dmpa...@email.com (formerly @clark.net, @ematic.com)
Erik Max Francis
> Erik Max Francis wrote:
>> Sylvia Else wrote:
>>> So the reasonable assumption is that it's in an orbit that allows it
>>> to be at 2000km at one point in time, and 800km at another.
>>
>> That would be my guess, too; it sounds asking what the speed of the
>> satellite will be if it has a perigee of 800 km altitude and an apogee
>> of 2000 km altitude. It's a bit glib and not terribly clear, though.
>> I agree with the others that it's not a very useful question,
>> especially for high school students.
>
> speed.
That seems in direct contradiction to what you just said above
It starts with an altitude of 2000 km, is hit by something, and ends up
at 800 km altitude. There's nothing about it ending up in a 800 km
_circular_ orbit.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis
Only the ephemeral is of lasting value.
-- Ionesco