Can an Astronaut Orbit the Space Shuttle?

I saw this question on the Internet. The professional astronomer replied that the answer is “no”, because the Earth’s gravitational field overwhelms the attraction between the astronaut and the shuttle. She mentioned that the shuttle’s Hill sphere was too small for orbiting to occur. I’d never heard of the Hill sphere, so a few hours on Google were utilized to investigate this concept.

We all know that the famous gravitational three-body problem has no explicit solution. Of course there are ways of solving it to any desired level of accuracy with computer techniques. In fact, the Hill sphere is often used as an analytic way of dealing with some aspects of this issue.

The Hill sphere is named for the nineteenth century American celestial mechanics expert, G. H. Hill (1838 – 1914). He developed a formulation to approximate the gravitational sphere of influence of a minor body orbiting a major body i.e. the shuttle orbiting the Earth or the Earth orbiting the Sun:

where

r_{H} is the radius of the Hill sphere surrounding the minor body

a is the orbital distance of the minor body from the major body

m is the mass of the minor body

M is the mass of the major body

To illustrate, we can calculate
the Hill sphere for the Earth. Here, the distance from
the Sun, a = 149.6 million km, the mass of Earth, m = 5.97x10^{24} kg, and
the Sun’s mass, M = 1.99x10^{30 }kg. The Hill
sphere for Earth is then about 1.5 million km (0.01 AU). That
implies that a gravitational influence sphere of 1.5 million km radius
surrounds Earth. Our Moon is approximately 370,000 km
from Earth, about 25% within Earth’s Hill sphere. This
means the Moon is quite safely contained within Earth’s gravitational field. Because of its larger mass and semi-major axis, Jupiter’s
Hill sphere radius is huge, 0.35 AU. This explains its
large satellite count and its affect on comet orbits.

Now to the question of an
astronaut orbiting the space shuttle. Let’s use the
liftoff mass of the shuttle, m = 10,600 kg (weight 104,000 kg/9.8m/s^{2})
versus the Earth’s mass, M = 5.97x10^{24} kg. Assume
the shuttle orbits at 300 km above the Earth’s surface. Then
its orbital distance, a = 300 km + the Earth’s radius (6367 km) is 6667 km. The Hill sphere for the shuttle would then be only about
0.56 m. That means the shuttle’s gravitational sphere
of influence is less than two feet around its center of gravity! Obviously an astronaut outside the shuttle’s exterior is
an order of magnitude beyond its effective gravitational influence, so no
shuttle orbiting can occur.

If a spaceship were located well
outside the gravitational influence of the Sun, in deep space, an astronaut *could* orbit the vehicle. Remember that gravitational attraction, while declining as
the square of the distance, continues to infinity. This
means that orbiting a spaceship is feasible only for massive vessels that are
very remote from interfering stars and planets. Several
asteroids have already been found to have satellites. The
Hill sphere formulation is a convenient celestial mechanics method to assess
orbiting capability.

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Maintained by Ray Shapp Page last updated 11/27/2003 |