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help - gravity problem



 
 
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  #1  
Old June 8th 09, 02:56 AM posted to sci.space.tech
dotcom
external usenet poster
 
Posts: 3
Default help - gravity problem

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?

  #2  
Old June 8th 09, 11:47 AM posted to sci.space.tech
Joe Pfeiffer
external usenet poster
 
Posts: 23
Default help - gravity problem

dotcom writes:

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).

  #3  
Old June 8th 09, 07:40 PM posted to sci.space.tech
Ken S. Tucker
external usenet poster
 
Posts: 740
Default help - gravity problem

On Jun 7, 6:56 pm, dotcom 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) 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

  #4  
Old June 8th 09, 10:49 PM posted to sci.space.tech
Robert Heller
external usenet poster
 
Posts: 17
Default help - gravity problem

At Mon, 8 Jun 2009 14:40:47 EDT "Ken S. Tucker" wrote:


On Jun 7, 6:56 pm, dotcom 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) 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


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
Deepwoods Software -- Download the Model Railroad System
http://www.deepsoft.com/ -- Binaries for Linux and MS-Windows
-- http://www.deepsoft.com/ModelRailroadSystem/


  #5  
Old June 8th 09, 10:53 PM posted to sci.space.tech
dotcom
external usenet poster
 
Posts: 3
Default help - gravity problem

On Jun 8, 11:56 am, dotcom 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) 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?



OK , I think I have the answer now. it was to do with the folly of
simply adding velocities. if we convert teh original speed to kinetic
energy , add the kinetic energy that is gained by the loss of
potential energy as the ht reduced from 2000 to 800 km then convert
the total kinetic energy back to speed we get the "right?" answer. I
guess what the question meant was that even though the satellite had
the correct speed for a circular orbit at 2000km , if that speed was
not in the right direction , ie perpendicular to a line towards the
center of the earth, then in fact it would be an elliptical orbit that
may come down to 800km ? I think!



  #6  
Old June 8th 09, 11:45 PM posted to sci.space.tech
Dr J R Stockton[_33_]
external usenet poster
 
Posts: 5
Default help - gravity problem

In sci.space.tech message be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000v
ba.googlegroups.com, Sun, 7 Jun 2009 21:56:02, dotcom
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.

--
(c) John Stockton, nr London, UK. Turnpike 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 Mail News.

  #7  
Old June 9th 09, 11:09 AM posted to sci.space.tech
Ken S. Tucker
external usenet poster
 
Posts: 740
Default help - gravity problem

On Jun 8, 2:49 pm, Robert Heller wrote:
At Mon, 8 Jun 2009 14:40:47 EDT "Ken S. Tucker" wrote:


On Jun 7, 6:56 pm, dotcom 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) 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


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.


Here's a ref on circular orbital speed,
http://ceres.hsc.edu/homepages/class...ics/sec10.html
and I'll suggest an examination of lunar recession,
http://www.astronomy.ohio-state.edu/...it4/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

  #8  
Old June 9th 09, 11:09 AM posted to sci.space.tech
dotcom
external usenet poster
 
Posts: 3
Default help - gravity problem

On Jun 9, 8:45 am, Dr J R Stockton
wrote:
In sci.space.tech message be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000v
ba.googlegroups.com, Sun, 7 Jun 2009 21:56:02, dotcom
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..
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 seeme

d
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.

--
(c) John Stockton, nr London, UK. Turnpike

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

il News.- Hide quoted text -

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

  #9  
Old June 11th 09, 08:32 PM posted to sci.space.tech
Ken S. Tucker
external usenet poster
 
Posts: 740
Default help - gravity problem

On Jun 9, 3:09 am, dotcom wrote:
On Jun 9, 8:45 am, Dr J R Stockton
wrote:

In sci.space.tech message be0aecfa-e255-449f-8846-1b5c2d10216f@n4g2000v
ba.googlegroups.com, Sun, 7 Jun 2009 21:56:02, dotcom
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..
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


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 seeme

d
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.


--
(c) John Stockton, nr London, UK. Turnpike

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


il News.- Hide quoted text -

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

  #10  
Old June 12th 09, 09:55 PM posted to sci.space.tech
Steve Willner
external usenet poster
 
Posts: 1,172
Default help - gravity problem

In article ,
dotcom 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
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
email may be sent to your ISP.)

 




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