# Path Independence

• August 15th 2010, 07:27 PM
larryboi7
Path Independence
The mass of the earth is approximately 6 x 10^27 g and that of the sun is 330,000 times as much. The gravitational constant is 6.7 x 10^-8 m^3/s^2 g. The distance of the earth from the sun is about 1.5 x 10^12 cm. Compute, approximately, the work necessary to increase the distance of the earth from the sun by 1 cm.
• August 15th 2010, 07:46 PM
mr fantastic
Quote:

Originally Posted by larryboi7
The mass of the earth is approximately 6 x 10^27 g and that of the sun is 330,000 times as much. The gravitational constant is 6.7 x 10^-8 m^3/s^2 g. The distance of the earth from the sun is about 1.5 x 10^12 cm. Compute, approximately, the work necessary to increase the distance of the earth from the sun by 1 cm.

Think about the area under the force versus distance graph for the Earth.
• August 16th 2010, 04:20 AM
HallsofIvy
The point of "path independence" here is that because the force is conservative, the integration along a path on which the earth is "moved" one cm more from the sun (which, of course, it does regularly in its orbit) is the same as the "anti-derivative" evaluated at $r_0$ and $r_0+ 1$ where $r_0$ is the initial distance from the sun to the earth in cm.

Since the force is given by $(GmM)r^{-2}$ (G is the "universal graviational constant", m is the mass of the earth, and M is the mass of the sun), what is the anti-derivative of that?