Ok, then how much mass would be required for them to attract each other? Also, if an astronaut could remain motionless, would the gravitational force from the space station pull him in? Mike
--- Olhomorto@aol.com wrote:
[I]f an astronaut could remain motionless, would the gravitational force from the space station pull him in?
I hope your're not trying to con someone into doing your physics homework for you. -:) Even so, it's a good question. Let's look at a simplified model, assume that - - the pressure of the light of the Sun and small number of particles in space at the ISS's altitude have no effect, and, - the centrifugal force from the speed of the ISS and the astronaut cancels out Earth's gravity. So all we have is ISS's and the astronaut's masses acting on each other. A NASA M-suit suit has a mass of about 45kg. To make the math easy, let's say the astronaut weights 155kg and the total mass of the astronaut and the suit is 200kg. http://www.astronautix.com/craftfam/spasuits.htm The ISS has a mass of 183,283 kg. http://en.wikipedia.org/wiki/International_Space_Station Look at some of the equations about gravity and acceleration on these two webpages. There're complicated but we'll work through them. http://en.wikipedia.org/wiki/Gravity http://en.wikipedia.org/wiki/Gravitational_constant A NASA M-suit has enough air to sustain an astronaut for about 7 hours. http://www.hq.nasa.gov/osf/4 Let's say Astronaut Leslie-Do-Wrong steps out the ISS air-lock and suddenly finds him or herself motionless floating about the length of a football field - 100 meters from the ISS. What happens? Since he or she is floating in the gravitational field of the 183,283 kg space station, he or she starts to fall back towards the ISS, just like a ball dropped from a second-story of house falls to back to the Earth. But because the ISS is so much smaller than the Earth - the speed at which Astronaut Leslie-Do-Wrong falls back to the ISS is much smaller. 1) How fast does the astronaut "fall" back to the space station? The rate he or she accelerates back to the ISS is mathematically described in the equation: acceleration = ( G * m_ISS ) / r^2 where - G is the gravitational constant - 6.6742E-11 m^3 s^-2 kg^-1 m_ISS is the mass of the space station or 183283 kilograms and r is the distance in meters between astronaut and the space station, 100 meters in our example. This works out to a really small number. A fast walking pace on Earth is about 1 meter a second. Astronaut Leslie-Do-Wrong will start accelerating back to the ISS at the whopping speed of - 0.00000000122327 meters a second or 1.22327 * 10^-9 m/s. It's going to take him or her a while to "fall" back to the ISS. 2) What is the astronaut's velocity at any one second as he or she drifts back to the ISS? The instanteous velocity of Astronaut Leslie-Do-Wrong, as he or she drifts back to the ISS is described by the equation: v_instantaneous = acceleration * time (seconds) So to accelerate so that she or he is falling at a walking pace speed of 1 meter a second will only take Astronaut Leslie-Do-Wrong a mere 817,492,753 seconds. 3) How long does it take for the astronaut to drift back the ISS? The relationship between distance traveled, acceleration and time is described by the equation: d = 1/2 x acceleration * time^2 For Astronaut Leslie-Do-Wrong, we know that he or she is 100 meters from the ISS and is accelerating back at 0.00000000122327 meters a second. "t" for time works out so that it will take Astronaut Lesile-Do-Wrong 404,347 seconds or 112 hours to "fall" back to the ISS. Unfortunately, Astronaut Leslie-Do-Wrong only has 7 hours of breathable air. Somebody had better stick their head out the air-lock and through him or her a line. Playing with the numbers some more, what if Leslie was only floating 15.8 meters away from the ISS, instead of 100 meters? It would take Leslie 7 hours and 3 seconds to float back to the air-lock. If he or she holds their breath. They'll just make it back into the air-lock. I've thrown together a quick spreadsheet in Excel that you can grab at - http://members.csolutions.net/fisherka/astronote/astromath/ISSgravityq1.xls I'm no great physics or math wizard, so if someone else here wants to proof the numbers and post some corrections, I'd appreciate it. In conclusion, because the force of gravity from the space station extends out infinitely around the station, the astronaut would fall back to the station even if he or she was a million miles away - assuming there was nothing else acting on the station and the astronaut. But it might take forever for him or her to complete the trip, but eventually they'd arrive. - Canopus56 P.S. - The next chance to see the ISS fly overhead from SLC Utah is: Tuesday 23 August 2005 6h00m17s AM Appears 5h55m29s 1.8m az:311.9d NW Transit 6h00m17s 0.7m az: 27.2d NNE h:28.3d Disappears 6h05m05s 4.0m az:102.5d ESE 5h56m49.52s ISS Close to Vega, mag=0.0. Separation: 1.384d Position Angle: 102.1d Good low horizon track on star chart Nautical twilight: 5:41am Sunrise: 6:45am ____________________________________________________ Start your day with Yahoo! - make it your home page http://www.yahoo.com/r/hs
A correction for a bad link - --- Canopus56 <canopus56@yahoo.com> wrote:
A NASA M-suit has enough air to sustain an astronaut for about 7 hours. http://www.hq.nasa.gov/osf/4
- a working link should be - Question 4: "How long can you stay in a space suit?" http://www.hq.nasa.gov/osf/qanda.html __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
Not addressed at all is the really important question: How much bran did the astronaut have for breakfast? --- Canopus56 <canopus56@yahoo.com> wrote:
Question 4: "How long can you stay in a space suit?"
____________________________________________________ Start your day with Yahoo! - make it your home page http://www.yahoo.com/r/hs
In that case you would have to add a small propulsion factor... B) 73 de n7zi Gary "Why buy something for ten bucks when you can make it for a hundred." JR ----- Original Message ----- From: "Chuck Hards" <chuckhards@yahoo.com> To: "Utah Astronomy" <utah-astronomy@mailman.xmission.com> Sent: Monday, August 15, 2005 8:57 PM Subject: Re: [Utah-astronomy] Gravity
Not addressed at all is the really important question:
How much bran did the astronaut have for breakfast?
--- Canopus56 <canopus56@yahoo.com> wrote:
Question 4: "How long can you stay in a space suit?"
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--- Chuck Hards <chuckhards@yahoo.com> wrote:
Not addressed at all is the really important question: How much bran did the astronaut have for breakfast?
Keep on reading down to question 16 "What are some of the special features of a spacesuit?" -:) http://www.hq.nasa.gov/osf/qanda.html#16 __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
Nice example and description. Thanks for the great illustration. One minor correction... the units of acceleration should be meters per second squared (m/s^2 or m*s^-2). It makes the other units in the computation work out correctly. Cheers, Tyler -----Original Message----- From: utah-astronomy-bounces+tylerallred=earthlink.net@mailman.xmission.com [mailto:utah-astronomy-bounces+tylerallred=earthlink.net@mailman.xmission.co m] On Behalf Of Canopus56 Sent: Monday, August 15, 2005 6:25 PM To: Utah Astronomy Subject: Re: [Utah-astronomy] Gravity --- Olhomorto@aol.com wrote:
[I]f an astronaut could remain motionless, would the gravitational force from the space station pull him in?
I hope your're not trying to con someone into doing your physics homework for you. -:) Even so, it's a good question. Let's look at a simplified model, assume that - - the pressure of the light of the Sun and small number of particles in space at the ISS's altitude have no effect, and, - the centrifugal force from the speed of the ISS and the astronaut cancels out Earth's gravity. So all we have is ISS's and the astronaut's masses acting on each other. A NASA M-suit suit has a mass of about 45kg. To make the math easy, let's say the astronaut weights 155kg and the total mass of the astronaut and the suit is 200kg. http://www.astronautix.com/craftfam/spasuits.htm The ISS has a mass of 183,283 kg. http://en.wikipedia.org/wiki/International_Space_Station Look at some of the equations about gravity and acceleration on these two webpages. There're complicated but we'll work through them. http://en.wikipedia.org/wiki/Gravity http://en.wikipedia.org/wiki/Gravitational_constant A NASA M-suit has enough air to sustain an astronaut for about 7 hours. http://www.hq.nasa.gov/osf/4 Let's say Astronaut Leslie-Do-Wrong steps out the ISS air-lock and suddenly finds him or herself motionless floating about the length of a football field - 100 meters from the ISS. What happens? Since he or she is floating in the gravitational field of the 183,283 kg space station, he or she starts to fall back towards the ISS, just like a ball dropped from a second-story of house falls to back to the Earth. But because the ISS is so much smaller than the Earth - the speed at which Astronaut Leslie-Do-Wrong falls back to the ISS is much smaller. 1) How fast does the astronaut "fall" back to the space station? The rate he or she accelerates back to the ISS is mathematically described in the equation: acceleration = ( G * m_ISS ) / r^2 where - G is the gravitational constant - 6.6742E-11 m^3 s^-2 kg^-1 m_ISS is the mass of the space station or 183283 kilograms and r is the distance in meters between astronaut and the space station, 100 meters in our example. This works out to a really small number. A fast walking pace on Earth is about 1 meter a second. Astronaut Leslie-Do-Wrong will start accelerating back to the ISS at the whopping speed of - 0.00000000122327 meters a second or 1.22327 * 10^-9 m/s. It's going to take him or her a while to "fall" back to the ISS. 2) What is the astronaut's velocity at any one second as he or she drifts back to the ISS? The instanteous velocity of Astronaut Leslie-Do-Wrong, as he or she drifts back to the ISS is described by the equation: v_instantaneous = acceleration * time (seconds) So to accelerate so that she or he is falling at a walking pace speed of 1 meter a second will only take Astronaut Leslie-Do-Wrong a mere 817,492,753 seconds. 3) How long does it take for the astronaut to drift back the ISS? The relationship between distance traveled, acceleration and time is described by the equation: d = 1/2 x acceleration * time^2 For Astronaut Leslie-Do-Wrong, we know that he or she is 100 meters from the ISS and is accelerating back at 0.00000000122327 meters a second. "t" for time works out so that it will take Astronaut Lesile-Do-Wrong 404,347 seconds or 112 hours to "fall" back to the ISS. Unfortunately, Astronaut Leslie-Do-Wrong only has 7 hours of breathable air. Somebody had better stick their head out the air-lock and through him or her a line. Playing with the numbers some more, what if Leslie was only floating 15.8 meters away from the ISS, instead of 100 meters? It would take Leslie 7 hours and 3 seconds to float back to the air-lock. If he or she holds their breath. They'll just make it back into the air-lock. I've thrown together a quick spreadsheet in Excel that you can grab at - http://members.csolutions.net/fisherka/astronote/astromath/ISSgravityq1.xls I'm no great physics or math wizard, so if someone else here wants to proof the numbers and post some corrections, I'd appreciate it. In conclusion, because the force of gravity from the space station extends out infinitely around the station, the astronaut would fall back to the station even if he or she was a million miles away - assuming there was nothing else acting on the station and the astronaut. But it might take forever for him or her to complete the trip, but eventually they'd arrive. - Canopus56 P.S. - The next chance to see the ISS fly overhead from SLC Utah is: Tuesday 23 August 2005 6h00m17s AM Appears 5h55m29s 1.8m az:311.9d NW Transit 6h00m17s 0.7m az: 27.2d NNE h:28.3d Disappears 6h05m05s 4.0m az:102.5d ESE 5h56m49.52s ISS Close to Vega, mag=0.0. Separation: 1.384d Position Angle: 102.1d Good low horizon track on star chart Nautical twilight: 5:41am Sunrise: 6:45am ____________________________________________________ Start your day with Yahoo! - make it your home page http://www.yahoo.com/r/hs _______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://www.utahastronomy.com
Why the heck, spelled h.e.l.l., are you waisting your time at the AG's office? Quoting Canopus56 <canopus56@yahoo.com>:
--- Olhomorto@aol.com wrote:
[I]f an astronaut could remain motionless, would the gravitational force from the space station pull him in?
I hope your're not trying to con someone into doing your physics homework for you. -:) Even so, it's a good question. Let's look at a simplified model, assume that -
- the pressure of the light of the Sun and small number of particles in space at the ISS's altitude have no effect, and,
- the centrifugal force from the speed of the ISS and the astronaut cancels out Earth's gravity.
So all we have is ISS's and the astronaut's masses acting on each other.
A NASA M-suit suit has a mass of about 45kg. To make the math easy, let's say the astronaut weights 155kg and the total mass of the astronaut and the suit is 200kg. http://www.astronautix.com/craftfam/spasuits.htm
The ISS has a mass of 183,283 kg. http://en.wikipedia.org/wiki/International_Space_Station
Look at some of the equations about gravity and acceleration on these two webpages. There're complicated but we'll work through them. http://en.wikipedia.org/wiki/Gravity http://en.wikipedia.org/wiki/Gravitational_constant
A NASA M-suit has enough air to sustain an astronaut for about 7 hours. http://www.hq.nasa.gov/osf/4
Let's say Astronaut Leslie-Do-Wrong steps out the ISS air-lock and suddenly finds him or herself motionless floating about the length of a football field - 100 meters from the ISS. What happens?
Since he or she is floating in the gravitational field of the 183,283 kg space station, he or she starts to fall back towards the ISS, just like a ball dropped from a second-story of house falls to back to the Earth. But because the ISS is so much smaller than the Earth - the speed at which Astronaut Leslie-Do-Wrong falls back to the ISS is much smaller.
1) How fast does the astronaut "fall" back to the space station?
The rate he or she accelerates back to the ISS is mathematically described in the equation:
acceleration = ( G * m_ISS ) / r^2
where -
G is the gravitational constant - 6.6742E-11 m^3 s^-2 kg^-1
m_ISS is the mass of the space station or 183283 kilograms and
r is the distance in meters between astronaut and the space station, 100 meters in our example.
This works out to a really small number. A fast walking pace on Earth is about 1 meter a second.
Astronaut Leslie-Do-Wrong will start accelerating back to the ISS at the whopping speed of -
0.00000000122327 meters a second or 1.22327 * 10^-9 m/s.
It's going to take him or her a while to "fall" back to the ISS.
2) What is the astronaut's velocity at any one second as he or she drifts back to the ISS?
The instanteous velocity of Astronaut Leslie-Do-Wrong, as he or she drifts back to the ISS is described by the equation:
v_instantaneous = acceleration * time (seconds)
So to accelerate so that she or he is falling at a walking pace speed of 1 meter a second will only take Astronaut Leslie-Do-Wrong a mere 817,492,753 seconds.
3) How long does it take for the astronaut to drift back the ISS?
The relationship between distance traveled, acceleration and time is described by the equation:
d = 1/2 x acceleration * time^2
For Astronaut Leslie-Do-Wrong, we know that he or she is 100 meters from the ISS and is accelerating back at 0.00000000122327 meters a second.
"t" for time works out so that it will take Astronaut Lesile-Do-Wrong 404,347 seconds or 112 hours to "fall" back to the ISS.
Unfortunately, Astronaut Leslie-Do-Wrong only has 7 hours of breathable air. Somebody had better stick their head out the air-lock and through him or her a line.
Playing with the numbers some more, what if Leslie was only floating 15.8 meters away from the ISS, instead of 100 meters? It would take Leslie 7 hours and 3 seconds to float back to the air-lock. If he or she holds their breath. They'll just make it back into the air-lock.
I've thrown together a quick spreadsheet in Excel that you can grab at -
http://members.csolutions.net/fisherka/astronote/astromath/ISSgravityq1.xls
I'm no great physics or math wizard, so if someone else here wants to proof the numbers and post some corrections, I'd appreciate it.
In conclusion, because the force of gravity from the space station extends out infinitely around the station, the astronaut would fall back to the station even if he or she was a million miles away - assuming there was nothing else acting on the station and the astronaut. But it might take forever for him or her to complete the trip, but eventually they'd arrive.
- Canopus56
P.S. - The next chance to see the ISS fly overhead from SLC Utah is:
Tuesday 23 August 2005 6h00m17s AM Appears 5h55m29s 1.8m az:311.9d NW Transit 6h00m17s 0.7m az: 27.2d NNE h:28.3d Disappears 6h05m05s 4.0m az:102.5d ESE 5h56m49.52s ISS Close to Vega, mag=0.0. Separation: 1.384d Position Angle: 102.1d Good low horizon track on star chart Nautical twilight: 5:41am Sunrise: 6:45am
____________________________________________________ Start your day with Yahoo! - make it your home page http://www.yahoo.com/r/hs
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participants (6)
-
Canopus56 -
Chuck Hards -
diveboss＠xmission.com -
Gary Liptrot -
Olhomorto＠aol.com -
Tyler Allred