# Equations of motion

1. Jun 22, 2013

Suppose we have a star and a planet with radius vectors r1 and r2 respectively in a fixed inertial coordinate frame. Relative position of planet from sun is r = r2 - r1

Why is the gravitational pull felt by the planet equals to F = Gm1m2 / r^2 * ( -r/r ) ?

Therefore, F= Gm1m2/r^3

Secondly, we want to find the relative motion of the planet with respect to the star.

Why is it that we have to substract the equation of motion of the star from that of the planet ?

m1 = mass of star
m2 = mass of planet

m2a2 = -Gm1m2r/r^3 ------ 1st eq
m1a1 = Gm1m2/r^3 -------- 2nd eq

Finally, a = G(m1+m2)r/r3

2. Jun 22, 2013

### Staff: Mentor

That is just the way gravity works. The r in the numerator is a vector here.
That is wrong.

The relative distance is r = r2 - r1, as you wrote above. If you want to calculate the second time-derivative of that, the minus sign stays there.

3. Jun 24, 2013

OK

I did a mistake in typing : F= Gm1m2r/r^3

Thanks a lot :)

4. Jun 26, 2013

### BobG

The magnitude of the force is F = Gm1m2 / r^2. The part in parenthesis should be the radius vector (r2-r1) divided by the magnitude of the difference. It just gives a direction to the gravitational force vector.
Should include the radius vector in the numerator, same as above.

Both objects are orbiting their combined center of mass. You can often get away with saying the planet orbits the star because the combined center of mass is usually very near the center of mass of the star (if you add the Earth's mass to the Sun's mass, the total will still be very close to the Sun's mass), but the true situation is that both the planet and the star are orbiting their combined center of mass (with the resulting wobble of the star being one of the ways we detect planets around other stars).

5. Jul 14, 2013