Posted by John on January 01, 2000 at 01:20:27:
In Reply to: Speed of Hubble is 26 239 km/h posted by John on December 31, 1999 at 21:49:00:
Jim Frysinger, a professor of physics pointed out to me an error in the logic used to determine the speed of an object in orbit.
First, we look at the centripetal force on the object, in this case Hubble. It is calculated by F_c = mv^2/r.
Where F_c (F sub c) is the centipetal force in newtons (N)
m is the mass of Hubble in kilograms (kg)
v is Hubble's orbital velocity in metres per second (m/s)
r is the distance between hubbles centre of gravity of hubble and that of the earth in metres (m)
Second, we know that the attractive force between two masses is equal to the product of their masses times the universal gravitational contant (UVC) and inversely proportional to the square of the distances between them, or
F = GMm/r^2
where F is the force of attraction in newtons (N)
G is the UVC and is equal to 6.67 x 10^-11 Nm^2/kg^2
m is the mass of Hubble in kilograms (kg)
M is the mass of the earth equal to 5.98 x 10^24 kg
Letting F_c = F, and the equations above set equal to each other, the resulting formula is: v^2=GM/r or v=(GM/r)^-2
The radius of the earth is 6370 km. Hubble is 590 km above the earth, so Hubbles distance from the centre of the earth is 6960 km or 6 960 000 m.
Plugging the numbers in the equations, v then equals 7.57 km/s or 27 253 km/h.
Hubbles orbital path equals 2pi(6960) km, or 43 731 km. Dividing Hubbles velocity of 27 253 km/h into its orbital path of 43 731 km, yields and orbital period of 1.60 h or 96.3 min.
So, now we all know Hubbles true speed.