what is the mathematical equation of an orbit or falling down?

Answer 1

Assuming the orbiting object's mass is neglegible compared to the large gravitating mass.....Kepler's law of planetary motion is given by ..

R^3 / T^2 = GM / 4(pi)^2.....where T is orbital period, and R is the orbital radius (or semi-major axis), and M is the mass of the central object, G = Grav. constant.

If the object has no orbital velocity.....the equation reduces to the familiar acceleration of free fall....

a = GM / R^2

..

Answer 2

For a 6400 km radius Earth, and g force of 10 meter/second^2, if you shot your self straight from a cannon
to a distance of 1km then by the curvature of the Earth that means you would be √(6400^2 +1)-6400 or 7.83 e-5km or 7.83cm
above the horizon. Your time of flight falling through 7.83 cm would be √(.1566m/10m/s^2)
or 1/8 of a second. Approximating a cruise missile type orbit,
of 40000 km around the Earth, you would complete that in
1 hour, 23-1/2 minutes. 8km /sec. Compare that with rotation
of the Earth which corresponds to 40000km/86,400s or .463
km/sec at the Equator. So if Earth spun itself about 17 times
faster, we and everything would float weightless at the
Equator

Just wanted to edit my answer and add a couple of things.
If you use the centrifugal definition of acceleration V^2/R=a=g=10m/s^2,
V=√(10*6.4e6)=8km/s for a circular orbit
Since 6400km is an altitude of 30 km, you can calculate
intercontinental ballistic ranges for a variety of suborbital
velocities, resulting radial and tangential accelerations.

radial a = v^2/R - GM./R^2; G=6.67e-11, M=6.0e24
A missile horizontally inserted at 30km altitude with velocity
< 8km/s will range out by resultant of
a = V^2/R-10.
t=√(2*3e4/a)
RangeDown=V*t
If the missile was inserted to the horizontal entry point by
artillery shot, then double to get the ballistic range..

If the missile is boosted at the insertion point to > 8km/s
the missile enters an elliptical orbit. Perigee, or Ro, is the insertion point, or 6400 km (more later)

Because we are starting with velocity and perigee, the complex differential equations are reduced so one obtains the semimajor axis and eccentricity as a function of initial velocity and distance:

semimajor axis a=
R*(2-R*V^2*(GM)^-1)
eccentricity e=
√(M*a-(RV)^2/G)*(M*a)^-1)

Answer 3

In astronomy, Kepler's laws of planetary motion are the three mathematical laws that describe the motion of planets in the Solar System.
If you search wikipedia for Kepler's Laws or ellipse you can find the equations.

The equation for a falling body is d = 1/2 (gt^2) where d is the distance, g is the gravitational constant (on Earth its 9.8 m/s^2), and t is the time falling.

Answer 4

Orbits tend to be elliptical. Good orbits are designed such that the semi-major and semi-minor axes are nearly the same.

Look up "ellipse" to get the actual formula.