remember that this rocket did not escape earths gravity..... it went straight up and only escaped our air...not traveles at 221,ooo miles per hour to escape the gravity and orbit like the space shuttle does...
Uhhh....I think you have an extra 1 in that number. More like 18 to 22,000 MPH.
Here's a little math to figure it out if you'd like!
Mean orbital speed
For orbits with small eccentricity, the length of the orbit is close to that of a circular one, and the mean orbital speed can be approximated either from observations of the orbital period and the semimajor axis of its orbit, or from knowledge of the masses of the two bodies and the semimajor axis.[citation needed]
v_o \approx {2 \pi a \over T}v_o \approx \sqrt{\mu \over a}
where v is the orbital velocity, a is the length of the semimajor axis, T is the orbital period, and μ=GM is the standard gravitational parameter. Note that this is only an approximation that holds true when the orbiting body is of considerably lesser mass than the central one, and eccentricity is close to zero.
Taking into account the mass of the orbiting body,
v_o \approx \sqrt{G (m_1 + m_2) \over r}
where m1 is the mass of the orbiting body, m2 is the mass of the body being orbited, r is specifically the distance between the two bodies (which is the sum of the distances from each to the center of mass), and G is the gravitational constant. This is still a simplified version; it doesn't allow for elliptical orbits, but it does at least allow for bodies of similar masses.
When one of the masses is almost negligible compared to the other mass, as the case for Earth and Sun, one can approximate the previous formula to get:
v_o \approx \sqrt{\frac{GM}{r}}
or assuming r equal to the body's radius
v_o \approx \frac{v_e}{\sqrt{2}}
Where M is the (greater) mass around which this negligible mass or body is orbiting, and ve is the escape velocity.
For an object in an eccentric orbit orbiting a much larger body, the length of the orbit decreases with orbital eccentricity e, and is an ellipse. This can be used to obtain a more accurate estimate of the average orbital speed:
v_o = \frac{2\pi a}{T}\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 -\frac{5}{256}e^6 -\frac{175}{16384}e^8 - \dots \right] [2]
The mean orbital speed decreases with eccentricity.
Topic: Rocket science (Read 4574 times)
