Posted by
Hemos
on from the checking-things-out dept.
hubie writes "The NEAR spacecraft flew by the asteroid Eros at a closest approach of only 3 miles! Despite what the story says, that is much less than the altitude of a commuter aircraft. Stay tuned for some expected cool closeup shots."
Eros Photographed, Buggers Discovered!!!
by
Sir.Cracked
·
· Score: 5
In a leak from the Defense Department, the first Photos of Eros revealed a race of insectoid beings that insider's are calling Buggers. Speculation is flying as to how to deal with the threat, with one idea of gathering all the most brilliant children up and putting them on a space station to train as generals. More at 11.
If you don't know what I'm talking about, read Ender's Game
-- Where are we going, and why am I in this handbasket?
You are right about being able to orbit at any altitude, of course, friction of the atmosphere would become significant at lower altitudes.
You were only partially right when you said that the earth's gravity pulls at 9.8 m/(s^2), actually, according to Universal Gravity, two bodies will attract eachother with a force governed by the following equation:
F=(G M m)/(r^2)
Where F is the force of attraction, G is the gravitational constant (about 6.67*10^-11 m^3/(kg/s^2)), M and m are the masses, and r is the distance between the the two bodies' centers of gravity.
That is all well and good, but it dosen't tell us what the force of attraction for the earth is. To discover that, we have to use another equation, Newton's second law:
F= m a
Where F is force, m is mass, and a is (you gussed it) acceleration.
We can set F equal to F and yeald:
m a = (G M m) / r^2
We can cancel out the two small 'm's to get:
a = (G M) / r^2
We need a 'r' and a 'M', so I will now stipulate that the radius of the earth is about 6.38*10^6 m and that the mass of the earth is 5.98*10^24 kg. An abitious slashdoter could most likely find much more accurate figures, but these will fit our purposes. Consequently, we can find the acceleration due to gravity with:
a = (6.67*10^-11 * 5.98*10^24) / ((6.38*10^6)^2)
a = 9.799 m/(s^2)
I hope you can see that 'a' would be significantly less were an object higher (say, 500km) from the surface of the earth, as 'r' would increase:
a = (6.67*10^-11 * 5.98*10^24) / ((6.38*10^6 + 500*1000)^2)
a = 8.427 m/(s^2)
Back to the original question: How fast must an object to travel to orbit the earth at the surface? To answer that, we have to call apon yet another equation, that of centripical acceleration in circular motion:
a = (v^2) / r
We know that the centripical acceleration is the acceleration due to gravity, which we found to be 9.799 m/(s^2) at the surface of the earth, and we know 'r' to be the radius of the earth, so to find 'v' (velocity) we could have:
v = sqrt( a r )
v = sqrt( 9.799 * 6.38*10^6)
v = 7906.81 m/s
v = 7.907 km/s (*)
As you can see, this is significantly less than your figure of 30,000 km/s!
(*) Us Americans, who's brains have been destroyed by the English System, would get more sense out of 17,687 mi/hr.
--
Jordan Bettis
``Wherever you go, there's another stupid sigfile quote.''
Slashdot Mirror
Here's a number of pictures of it:
http://nssdc.gs fc. nasa.gov/planetary/mission/near/near_eros.html
Enjoy =)
---
In a leak from the Defense Department, the first Photos of Eros revealed a race of insectoid beings that insider's are calling Buggers. Speculation is flying as to how to deal with the threat, with one idea of gathering all the most brilliant children up and putting them on a space station to train as generals. More at 11.
If you don't know what I'm talking about, read Ender's Game
Where are we going, and why am I in this handbasket?
You are right about being able to orbit at any altitude, of course, friction of the atmosphere would become significant at lower altitudes.
You were only partially right when you said that the earth's gravity pulls at 9.8 m/(s^2), actually, according to Universal Gravity, two bodies will attract eachother with a force governed by the following equation:
F=(G M m)/(r^2)
Where F is the force of attraction, G is the gravitational constant (about 6.67*10^-11 m^3/(kg/s^2)), M and m are the masses, and r is the distance between the the two bodies' centers of gravity.
That is all well and good, but it dosen't tell us what the force of attraction for the earth is. To discover that, we have to use another equation, Newton's second law:
F= m a
Where F is force, m is mass, and a is (you gussed it) acceleration.
We can set F equal to F and yeald:
m a = (G M m) / r^2
We can cancel out the two small 'm's to get:
a = (G M) / r^2
We need a 'r' and a 'M', so I will now stipulate that the radius of the earth is about 6.38*10^6 m and that the mass of the earth is 5.98*10^24 kg. An abitious slashdoter could most likely find much more accurate figures, but these will fit our purposes. Consequently, we can find the acceleration due to gravity with:
a = (6.67*10^-11 * 5.98*10^24) / ((6.38*10^6)^2)
a = 9.799 m/(s^2)I hope you can see that 'a' would be significantly less were an object higher (say, 500km) from the surface of the earth, as 'r' would increase:
a = (6.67*10^-11 * 5.98*10^24) / ((6.38*10^6 + 500*1000)^2)
a = 8.427 m/(s^2)Back to the original question: How fast must an object to travel to orbit the earth at the surface? To answer that, we have to call apon yet another equation, that of centripical acceleration in circular motion:
a = (v^2) / r
We know that the centripical acceleration is the acceleration due to gravity, which we found to be 9.799 m/(s^2) at the surface of the earth, and we know 'r' to be the radius of the earth, so to find 'v' (velocity) we could have:
v = sqrt( a r )
v = sqrt( 9.799 * 6.38*10^6) v = 7906.81 m/s v = 7.907 km/s (*)As you can see, this is significantly less than your figure of 30,000 km/s!
(*) Us Americans, who's brains have been destroyed by the English System, would get more sense out of 17,687 mi/hr.
Jordan Bettis
``Wherever you go, there's another stupid sigfile quote.''