# The Physics, Astronomy and Mathematics of the Solar System

## Mathematics at the Frontier of Astronomy

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## Orbital Motion

From the orbital motion of two gravitating bodies information about the mass of the bodies can be calculated. If the two masses orbiting under the influence of the same gravitational force, they will have orbits about a common point. This point is called the center of mass.

The orbits will always be of one of three kind of curve an ellipse, parabolic, or hyperbolic.

*
*

### The center of mass

The center of mass is a point directly between the two masses. A straight line can be drawn to connect both masses. Let
*
m
*
represents the smaller mass and
*
M
*
represents the larger mass, then the distance of
*
M
*
from the center of mass represented by
*
R
*
and distance of
*
m
*
from the center of mass is represented by
*
r
*
. This distance can be expressed as

*
MR = mr.
*

For very large objects like the planets the center of mass can be located at the geometric center of the body. For example in the Earth- Moon orbital system, the center of mass is inside the Earth but not at the geometrical center. Both the Earth and the Moon orbit about a common center of mass, but the orbit of the Earth is so small because of its mass when compared to the Moon, it is said that the Moon orbits the Earth. The center of mass of the Earth - Moon system is located in the center of the Earth. The Earth and the Moon orbit around a common center of mass. For two bodies of considerable differences in masses the center of mass is closer to the heavier body. The concept of the center of mass can be extended to systems of a number of mass points.

The larger mass
*
M
*
has a gravitational attraction on the smaller mass
*
m
*
. Therefore

Newton's second law for gravitational for masses can be applied. The law states that the acceleration of an object is proportional to the net outside force acting on the object.

In an equation this is expressed as
*
F = ma
*
. Given that
*
F
*
= GMm/r^2 then GMm/(r+R)^2 = m(v^2/r). When
*
v
*
= (2 Pi ^2 * r )/ T. Replacing
*
v
*
with
*
2Pi * r
*
we have GM / (r+R)^2 = (4 Pi ^2 * r)/T^2 this equation becomes
*
T
^{
2
}
*
= (4 Pi ^2 * d^3) / [G(m+M)]

*d*

*=(r + R)*and

*T = period*.

For a small mass orbiting a much larger mass, the equation can be applied to estimate the mass of the heavier body if the orbital data of the lighter mass is known.

Problem: Estimate the mass of the sun given the following information

*
Let M
*
= Mass of the Sun

*
Let m
*
= Mass of the Earth

*
d
*
=
*
150,000,000 km
^{
14
}
*

*
T =
*
1 year
*
*

Converting to mks and substituting in the equation, the mass of the Sun

*
M
*
*
= 2. 00 × 10
^{
30
}
kg.
*
The mass of the Sun is used as the unit for expressing the mass of other stars and other large bodies. The symbol used for the mass of the Sun is M.

**
**