# Weighing The Sun

We can use Newtonian mechanics to calculate some useful formula relating the properties of planetary orbits to the mass of the Sun. Again for simplicity, let’s assume that the orbits are circular (not a bad approximation in most cases, and Newton’s laws easily extend to cover the more general case of eccentric orbits). Consider a planet of mass

*m*moving at speed*v*in an orbit of radius*r*around the Sun, of mass M. The planet’s acceleration (see*More Precisely 2-2*) isso, by Newton’s second law, the force required to keep it in orbit is

Comparing this with the gravitational force due to the Sun, it follows that

so the circular orbit speed is

Dividing this speed into the circumference of the orbit (2π

*r*), we obtain a form of Kepler’s third law (equivalent to the formula presented in the text):where

*P*= 2π*r*/*v*is the orbital period.Now we turn the problem around. Because we have measured G in the laboratory on Earth and because we know the length of a year and the size of the astronomical unit, we can use Newtonian mechanics to

*weigh*the Sun. Rearranging the above equation to readand substituting the known values of

*v*= 30 km/s,*r*= 1 A.U. = 1.5 x 10^{11}m, and*G*= 6.7 x10^{-11}N m^{2}/kg^{2}, we calculate the mass of the Sun to be 2.0 x 10^{30}kg—an enormous mass by terrestrial standards. Similarly, knowing the distance to the Moon and the length of the (sidereal) month, we can measure the mass of Earth to be 6.0 x 10^{24}kg.In fact, this is how basically

*all*masses are measured in astronomy. Because we can’t just go out and attach a scale to an astronomical object when we need to know its mass, we must look for its gravitational influence on something else. This principle applies to planets, stars, galaxies, and even clusters of galaxies—very different objects but all subject to the same physical laws.
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