some orbital mechanics

Are orbiting objects weightless?

No! Not if you believe the universal law of gravitation. In vector form it says that the gravitational force F between a massive object (mass M) and a less massive object (mass m), that are separated by a center-to-center distance r, is:

G M m  r  F = -   , (Eq. 1) r2  r  where G is the gravitation constant of 6.67 x 10-11 N m2 kg-2, and r is the vector (of magnitude r) connecting the centers of mass of M and m.

For an orbiting satellite, r is more conveniently written as: r = rearth + z, with z being the satellite’s altitude above mean sea level. At sea level, clearly r = rearth.

What is the ratio of the magnitudes of the gravity forces at sea level and at, say, a 300-km altitude space shuttle? Now the masses of the earth and shuttle (M and m, respectively) and G cancel, leaving:

|F(300 km)|  rearth  2  6371 km 2 =   =   = 0.912, |F(0 km)|  rearth + z  6671 km which is less than a 10% reduction in the acceleration of gravity. How can we reconcile this number with images of “weightless” astronauts tumbling about their quarters?

ANSWER: We confuse sustained free fall (very definitely a response to the gravity of the earth) with the absence of gravity. Ironically, free fall requires a strong gravitational field, but true weightlessness requires the absence of gravitational acceleration.

What does occur in an orbiting satellite (whether the moon, Landsat, or the space shuttle)? The satellite’s very high forward velocity is combined with its gravity- induced downward velocity to form a curve that doesn’t intersect the earth or the lower atmosphere.

In other words, gravity is a prerequisite for orbit, not a force to be escaped from in orbit. Even moon- or planet-bound vehicles are not weightless, strictly speaking. They simply have a high enough forward velocity that their “orbits” are no longer closed curves about the earth.

So everywhere that spacecraft travel, the earth’s gravity affects them (as described in Eq. 1), although it quickly becomes negligible. -2- calculating orbital parameters

What linear velocity is required to maintain a circular orbit at altitude z? Here we assume no air resistance, although clearly that affects low-orbiting satellites.

Start by noting that, in time interval t, a projectile with a constant linear velocity v0 covers a horizontal distance d = v0t. During the same time it falls a vertical 1 2 distance z, where z = 2 g t assumes a constant gravitational acceleration g.

d = v0t 2 } z = 1/2 gt

earth

Because d is a horizontal, tangential distance, we can include below it in a cross- sectional view of the projectile’s path above the spherical earth’s surface.

d2 + r2 = (r + z)2 leads to z d2 + r2 = r2 + 2rz + z2 or d d2 ≈ 2rz if z « r, which it is for low earth orbit

earth On a spherical earth, traveling a tangent distance d results in the earth’s curved surface falling away a r vertical distance z beneath you. From the 2 2 2 Pythagorean Theorem, d + rearth = (rearth + z) . r 2 If z << rearth , the z term is negligible, and 2 expanding the above equation yields: d = 2 rearth z.

If in time t gravity makes the projectile fall the same distance z 1 2 (where z = 2 g t ), then we can then equate the two vertical displacements (one for the earth’s surface, and one for the falling projectile).

SO431 — some orbital mechanics -3- 2 1 2 2 Then we get d = 2 rearth z = 2 rearth ( 2 g t ) = rearth g t or 1/2 d = t (rearth g) . For uniform velocity v0, this becomes: d 1/2 v0 {= t } = (rearth g) , (Eq. 2)

-2 the velocity required for orbit near the earth’s surface. For g = 9.8 m sec and rearth 6 -1 = 6.371 x 10 m, v0 ~ 7900 m sec , which is ~ 6 times faster than a fast rifle bullet.

A more sophisticated orbital velocity equation also accounts for the orbit’s radius (we’ll stick to circular orbits here). Starting with an equation that considers the acceleration of the earth’s gravity and the satellite’s kinetic energy, we can show that for a satellite orbiting the earth:

2  g rearth  1/2 v =   . (Eq. 3) 0  rearth + z

Note that the required orbital velocity decreases as z increases. Then why not put all satellites in high-altitude orbits?

What is the time T required for a single orbit? By definition, a satellite’s 2π v angular velocity ω = , where ω is in radians/sec. However, we also have ω = , T rsat where rsat = rearth + z. In other words, the satellite’s linear velocity v is its angular velocity times its orbital radius.

In orbit, a satellite’s centripetal acceleration comes from the gravitational acceleration between it and the earth, so that:

2 2 2 m v G M m v G M rearth 2 = 2 ⇒ = 2 = g 2 (because G M = g rearth ). rsat rsat rsat rsat rsat

2 2 2 2 2 rearth Since v = ω rsat , this last equation becomes ω = g 3 . Given that rsat π 2 3 2π 2 (2 ) rsat ω = , this equation can in turn be rewritten as: T = 2 or T g rearth

3/2 2π rsat T = . (Eq. 4) g1/2 rearth

SO431 — some orbital mechanics