Newton's Gravitational Law [closed]

Question

Is this correct..?

Gravitational constant, $G = 6.67 × 10^{-11} Nm^2kg^{-2}$

Mass of Sun, $M = 1.99 × 10^{30} kg$

Mass of Earth, $m= 6.02 × 10^{24}kg$

Distance between Sun and Earth, $R= 1.5 × 10^{11}m$

Hence, the Force acting on Earth by Sun, $F = \frac{GMm}{R^2} = 3.39 × 10^{22}$ $N$

Answer 1

Well, the numbers you gave plug in to get $F = 3.55 \times 10^{22}\ \mathrm{N}$. That large a discrepancy probably with what you quote as the force probably comes from doing the calculation in stages in rounding at each step. Always plug in numbers in a single step at the very end.

Also, I see $M_\oplus = 5.97 \times 10^{24}\ \mathrm{kg}$ in several sources. You might want to use that instead.

In fact, there's an important lesson hidden away beneath this simple calculation. You may have heard people say $G$ is one of the least precisely measured constants. The thing is, it's hard to design experiments to measure $G$. Conceptually, you want to take two known masses, separated them with a known distance, and measure the force of attraction, just as Cavendish did The problem is that this force will be exceedingly small for any two reasonable masses. You might want to use the Earth as one of the masses, but that begs the question, since you need $G$ to find the mass of the Earth. (A clever way to try to avoid this was the Schiehallion experiment, which was able to say something about the average density of the Earth.)

As it happens, studying the orbits of the planets over very long times gives exquisite precision for a number of parameters, but these are not necessarily the parameters you're looking for. The heliocentric gravitational constant, $G M_\odot$, is known with several orders of magnitude better precision than either $G$ or $M_\odot$ alone.