Problem
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
[MATH]F_g=G∗\frac{m1∗m2}{r^2}[/MATH][MATH]F=ma[/MATH][MATH]F_{cent}=m∗\frac{v^2}{r}[/MATH]
Thoughts
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
[MATH]F=G∗\frac{m1∗m2}{r^2}[/MATH][MATH]G=6.67∗10^{−11}\frac{Nm^2}{kg^2}[/MATH][MATH]m1=5.97∗10^{24}kg∗.90[/MATH][MATH]m2=5.97∗10^{24}kg∗.50[/MATH][MATH]r_{min}=3.84∗10^6m [/MATH][MATH]r_{max}=1.158∗10^7m [/MATH]
Plugging these values in I get [MATH]F_{min}=7.25∗10^{25}N[/MATH] and [MATH]F_{max}=1.81∗10^{25}N[/MATH]
[MATH]F_{min}[/MATH] is at minimum distance and [MATH]F_{max}[/MATH] is at maximum distance
That is a pretty strong gravitational force between the 2 masses. I may be off by a few orders of magnitude but it is still quite a strong force.
Gravitational acceleration of each mass at minimum distance:
[MATH]a1_{min}=13.49\frac{m}{s^2}[/MATH][MATH]a2_{min}=24.29\frac{m}{s^2}[/MATH]
Gravitational acceleration of each mass at maximum distance:
[MATH]a1_{max}=3.37\frac{m}{s^2}[/MATH][MATH]a2_{max}=6.06\frac{m}{s^2}[/MATH]
Where [MATH]a1[/MATH] is the acceleration of [MATH]m1[/MATH] and [MATH]a2[/MATH] is the acceleration of [MATH]m2[/MATH].
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:
[MATH]F_c=F_g[/MATH][MATH]F_c=m∗\frac{v^2}{r}[/MATH][MATH]F_c∗r=m∗v^2[/MATH][MATH]\frac{F_c∗r}{m}=v^2[/MATH][MATH]\sqrt{\frac{F_c∗r}{m}}=v[/MATH]
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 [MATH]m1[/MATH]:
[MATH]v_{peri}=7198.23\frac{m}{s}[/MATH][MATH]v_{aph}=5394.95\frac{m}{s}[/MATH]
And for [MATH]m2[/MATH]:
[MATH]v_{peri}=9657.45\frac{m}{s}[/MATH][MATH]v_{aph}=7238.09\frac{m}{s}[/MATH]
But as it turns out, this is the wrong velocity. This is the minimum velocity for a circular orbit, not the orbital velocity of the object. So the algebraic approach is out and I have to use calculus to go any further than knowing the accelerations and gravitational force at 2 points for each mass.
But how am I supposed to go from knowing the force and acceleration at 2 points to knowing the entire graph of the orbital velocity? Once I know the entire graph, calculating the period of the orbit should be easy just looking at the graph or doing some algebra.
Since the acceleration changes, the minimum number of derivative operators needed to arrive at a constant value is 3. But there may be no point where the derivative is 0 everywhere. If there is no such point where the xth derivative = 0 everywhere, then I can't simply do a chain of antiderivatives to arrive at the velocity graph.
So how can I go from knowing the force and acceleration at 2 points for each mass to knowing the entire velocity graph and thus the period of the binary orbit?
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
[MATH]F_g=G∗\frac{m1∗m2}{r^2}[/MATH][MATH]F=ma[/MATH][MATH]F_{cent}=m∗\frac{v^2}{r}[/MATH]
Thoughts
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
[MATH]F=G∗\frac{m1∗m2}{r^2}[/MATH][MATH]G=6.67∗10^{−11}\frac{Nm^2}{kg^2}[/MATH][MATH]m1=5.97∗10^{24}kg∗.90[/MATH][MATH]m2=5.97∗10^{24}kg∗.50[/MATH][MATH]r_{min}=3.84∗10^6m [/MATH][MATH]r_{max}=1.158∗10^7m [/MATH]
Plugging these values in I get [MATH]F_{min}=7.25∗10^{25}N[/MATH] and [MATH]F_{max}=1.81∗10^{25}N[/MATH]
[MATH]F_{min}[/MATH] is at minimum distance and [MATH]F_{max}[/MATH] is at maximum distance
That is a pretty strong gravitational force between the 2 masses. I may be off by a few orders of magnitude but it is still quite a strong force.
Gravitational acceleration of each mass at minimum distance:
[MATH]a1_{min}=13.49\frac{m}{s^2}[/MATH][MATH]a2_{min}=24.29\frac{m}{s^2}[/MATH]
Gravitational acceleration of each mass at maximum distance:
[MATH]a1_{max}=3.37\frac{m}{s^2}[/MATH][MATH]a2_{max}=6.06\frac{m}{s^2}[/MATH]
Where [MATH]a1[/MATH] is the acceleration of [MATH]m1[/MATH] and [MATH]a2[/MATH] is the acceleration of [MATH]m2[/MATH].
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:
[MATH]F_c=F_g[/MATH][MATH]F_c=m∗\frac{v^2}{r}[/MATH][MATH]F_c∗r=m∗v^2[/MATH][MATH]\frac{F_c∗r}{m}=v^2[/MATH][MATH]\sqrt{\frac{F_c∗r}{m}}=v[/MATH]
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 [MATH]m1[/MATH]:
[MATH]v_{peri}=7198.23\frac{m}{s}[/MATH][MATH]v_{aph}=5394.95\frac{m}{s}[/MATH]
And for [MATH]m2[/MATH]:
[MATH]v_{peri}=9657.45\frac{m}{s}[/MATH][MATH]v_{aph}=7238.09\frac{m}{s}[/MATH]
But as it turns out, this is the wrong velocity. This is the minimum velocity for a circular orbit, not the orbital velocity of the object. So the algebraic approach is out and I have to use calculus to go any further than knowing the accelerations and gravitational force at 2 points for each mass.
But how am I supposed to go from knowing the force and acceleration at 2 points to knowing the entire graph of the orbital velocity? Once I know the entire graph, calculating the period of the orbit should be easy just looking at the graph or doing some algebra.
Since the acceleration changes, the minimum number of derivative operators needed to arrive at a constant value is 3. But there may be no point where the derivative is 0 everywhere. If there is no such point where the xth derivative = 0 everywhere, then I can't simply do a chain of antiderivatives to arrive at the velocity graph.
So how can I go from knowing the force and acceleration at 2 points for each mass to knowing the entire velocity graph and thus the period of the binary orbit?
