What is the period of this binary orbit?

[caters]

Homework Statement

There are 2 masses, one with 90% the mass of earth and one with 50% the mass of earth. Let's call the bigger one m1 and the smaller one m2. m1 has a radius of 95% that of earth and m2 has a radius of 40% that of earth. These 2 masses are in a binary orbit. The minimum distance in this binary orbit is the earth-moon distance. The maximum distance in the binary orbit is 2.25 * the earth-moon distance. At minimum distance, the 2 masses are at their maximum velocity and at maximum distance, they are at their minimum velocity. Calculate the period of the binary orbit using Newton's laws of motion and the Law of Universal Gravitation. Use calculus if necessary.

Relevant Equations

$$F_g = G*\frac{m1*m2}{r^2}$$
$$F = ma$$
$$F_{cent} = \frac{m * v^2}{r}$$

Okay, so let's simplify things and look at the ideal case, where the 2 masses are the same. Well, this gives us a sine wave pattern to the orbital velocity. Consequently, the acceleration would follow a cosine wave since the derivative of velocity is acceleration and the derivative of sine is cosine. Since this acceleration is from the gravitational force of the 2 masses, the gravitational force would have to change in proportion to cos(t). Now, getting to the real problem.

Figuring out gravitational force:
$$F = G * \frac{m1*m2}{r^2}$$
$$G = 6.67*10^{-11} \frac{Nm^2}{kg^2}$$
$$m1 = 5.97*10^{24} kg * .90$$
$$m2 = 5.97*10^{24} kg * .50$$
$$r_{min} = 3.84*10^6 m$$
$$r_{max} = 1.158 * 10^7m$$
Plugging these values in I get $F_{min} = 7.25 * 10^{25} N$ and $F_{max} = 1.81 * 10^{25} N$

$F_{min}$ is at minimum distance and $F_{max}$ is at maximum distance

That is a pretty strong gravitational force between the 2 masses.

Gravitational acceleration of each mass at minimum distance:
$$a1_{min} = 13.49 m/s^2$$
$$a2_{min} = 24.29 m/s^2$$

Gravitational acceleration of each mass at maximum distance:
$$a1_{max} = 3.37 m/s^2$$
$$a2_{max} = 6.06 m/s^2$$

Where $a1$ is the acceleration of $m1$ and $a2$ is the acceleration of $m2$.

If I equate the gravitational force, which is what I have calculated, with the centripetal force, I can get the velocity at minimum and maximum distance just doing some algebra. Here is what I get after doing the algebra:

$$F_c = F_g$$
$$F_c = \frac{m*v^2}{r}$$
$$F_c * r = m * v^2$$
$$\frac{F_c*r}{m} = v^2$$
$$\sqrt{\frac{F_c*r}{m}} = v$$

And then if I plug in the masses, gravitational force, and distance at aphelion(maximum distance and perihelion(minimum distance, I will get these velocites for $m1$:

$$v_{peri} = 7198.23 \frac{m}{s}$$
$$v_{aph} = 5394.95 \frac{m}{s}$$

And for $m2$:

$$v_{peri} = 9657.45 \frac{m}{s}$$
$$v_{aph} = 7238.09 \frac{m}{s}$$

But now, what do I do? I know the velocities of both masses at perihelion and aphelion as well as their accelerations at those points and the gravitational force at those points. But how am I supposed to go from knowing the velocities, accelerations, and the gravitational force at 2 points to figuring out the period of the binary orbit?

 

[osilmag]

Nice show of your work.

Orbital speed can also be calculated with $v=\sqrt{\frac{GM} {R}}$.

You might want to look up Kepler's 3rd Law for the period.

 

[osilmag]

Your velocity can also be set equal to $\frac {2\pi r} {T}$.

 

[Janus]

Problem Statement: There are 2 masses, one with 90% the mass of Earth and one with 50% the mass of earth. Let's call the bigger one m1 and the smaller one m2. m1 has a radius of 95% that of Earth and m2 has a radius of 40% that of earth. These 2 masses are in a binary orbit. The minimum distance in this binary orbit is the earth-moon distance. The maximum distance in the binary orbit is 2.25 * the earth-moon distance. At minimum distance, the 2 masses are at their maximum velocity and at maximum distance, they are at their minimum velocity. Calculate the period of the binary orbit using Newton's laws of motion and the Law of Universal Gravitation. Use calculus if necessary.
Relevant Equations: $$F_g = G*\frac{m1*m2}{r^2}$$
$$F = ma$$
$$F_{cent} = \frac{m * v^2}{r}$$

Okay, so let's simplify things and look at the ideal case, where the 2 masses are the same. Well, this gives us a sine wave pattern to the orbital velocity. Consequently, the acceleration would follow a cosine wave since the derivative of velocity is acceleration and the derivative of sine is cosine. Since this acceleration is from the gravitational force of the 2 masses, the gravitational force would have to change in proportion to cos(t). Now, getting to the real problem.

Figuring out gravitational force:
$$F = G * \frac{m1*m2}{r^2}$$
$$G = 6.67*10^{-11} \frac{Nm^2}{kg^2}$$
$$m1 = 5.97*10^{24} kg * .90$$
$$m2 = 5.97*10^{24} kg * .50$$
$$r_{min} = 3.84*10^6 m$$
$$r_{max} = 1.158 * 10^7m$$
Plugging these values in I get $F_{min} = 7.25 * 10^{25} N$ and $F_{max} = 1.81 * 10^{25} N$

Your answers are off by a few magnitudes. The Earth-Moon distance is 3.84e8 m, not 3.84e6m. Thus your answer is 10,000 times too large.

$F_{min}$ is at minimum distance and $F_{max}$ is at maximum distance

That is a pretty strong gravitational force between the 2 masses.

Gravitational acceleration of each mass at minimum distance:
$$a1_{min} = 13.49 m/s^2$$
$$a2_{min} = 24.29 m/s^2$$

Gravitational acceleration of each mass at maximum distance:
$$a1_{max} = 3.37 m/s^2$$
$$a2_{max} = 6.06 m/s^2$$

Where $a1$ is the acceleration of $m1$ and $a2$ is the acceleration of $m2$.

If I equate the gravitational force, which is what I have calculated, with the centripetal force, I can get the velocity at minimum and maximum distance just doing some algebra. Here is what I get after doing the algebra:

$$F_c = F_g$$
$$F_c = \frac{m*v^2}{r}$$
$$F_c * r = m * v^2$$
$$\frac{F_c*r}{m} = v^2$$
$$\sqrt{\frac{F_c*r}{m}} = v$$

First, I assume you meant F_g in this last equation. As written, it just simplifies to v=v. Secondly, from the answers below, I is apparent that you took r to equal the distance between the two bodies when working out F_c, and this would not be the case. The two bodies would be orbiting the barycenter between them, and it would be the distance between this and the body in question that should be used for r here. Thirdly, this will not give you the periapis and apoapis velocities ( Aphelion and perihelion refer to orbits around the Sun, just like apogee and apogee refer to Earth orbits, periapis an apoapis are the general terms for any orbit.). While Fg=Fc holds true at all points of a circular orbit, it does not do so for elliptical ones. So even when you correct for the erroneous value for the gravitational force between the bodies and properly account for the motion around the barycenter, the method above would not give you the correct maximum and minimum velocities of the bodies. What your method gives you is the orbital velocity needed to maintain a circular orbit at those distances.

And then if I plug in the masses, gravitational force, and distance at aphelion(maximum distance and perihelion(minimum distance, I will get these velocites for $m1$:

$$v_{peri} = 7198.23 \frac{m}{s}$$
$$v_{aph} = 5394.95 \frac{m}{s}$$

And for $m2$:

$$v_{peri} = 9657.45 \frac{m}{s}$$
$$v_{aph} = 7238.09 \frac{m}{s}$$

But now, what do I do? I know the velocities of both masses at perihelion and aphelion as well as their accelerations at those points and the gravitational force at those points. But how am I supposed to go from knowing the velocities, accelerations, and the gravitational force at 2 points to figuring out the period of the binary orbit?

Such a numerical approach isn't likely to give results. A better angle of attack is to work with the relevant equations themselves; using them, and the above suggested calculus to derive an equation for orbital period, into which you can plug in the given values to get your answer.