# Launch to Orbit, Part 2

See Part 1, where I looked at how increasing horizontal launch velocities could lead to orbit. Now I want to explore this mathematically. In the Feynman Lectures on Physics, Vol. I, Ch. 7, Section 7–4 “Newton’s law of gravitation,” Feynman looks at the velocity required to achieve (circular) orbit. He looks at how far a projectile would fall in one second, and then looks at how fast it would have to be traveling horizontally to clear the surface and maintain the same altitude. I’d like to expand on his approach and see if I can find a general expression for the velocity.

Problem: Use the standard projectile motion equations to derive an expression for the velocity required to achieve circular orbit.

When I say I want to use the standard projectile motion equations, I am referring to s = ½at2 + v0s0, which gives us the position s as a function of time t. This equation assumes a constant acceleration a. Of course the direction of acceleration will be changing, but if we assume a short enough time interval, we (hopefully) can assume that gravity — and therefore acceleration — will only be in the y direction.

Feynman provides a diagram, but while his approach is (of course) geometrically sound, it’s not the most intuitive way for me to approach the problem. I drew my own diagram:

In this diagram, there is a circle representing the orbit, centered at O. Point A represents the starting position of the projectile. After a time t, it will have moved to the right. If there were no gravity, it would continue in a straight line to point B. However, gravity will pull it down (again, we’re going to assume it is pulled down as in the negative y direction, not towards the center of the planet O, as that would be much more complicated). The amount it falls depends only on time: the x and y movement are independent. It will take a certain amount of time for it to fall the distance BC. If the projectile is launched with sufficient x velocity, in the time it takes for the projectile to fall that distance, it will reach a horizontal distance equal to AB and therefore will reach point C, on the requisite circle.

It was tough for me to conceptualize this, so I’m going to break this problem down into four steps. 1, find how far the projectile will fall in the y direction after a time t2, find how far in the x direction it must travel in that time to stay on the circle. 3, find the horizontal velocity needed to achieve the x distance in that time. 4, look at very small time intervals so that our assumptions hold (in particular, that the direction of “down” doesn’t change significantly.

# Step 1: Find how far the projectile falls in the y direction

We know that

$\displaystyle y = \frac{1}{2}a_yt^2 + v_{0,y}t + y_0$

It has no initial y velocity, so v0,y=0.

$\displaystyle y = \frac{1}{2}a_yt^2 + y_0$

The initial height is equal to the radius of orbit r.

$\displaystyle y = \frac{1}{2}a_yt^2 + r$

What about the acceleration? I don’t simply want to use −g, since g represents the acceleration due to gravity at Earth’s surface, and I want to have a general expression that will hold at varying distances from Earth (or whichever planet/star). The force due to gravitation is given by Newton:

$\displaystyle F = -G\frac{Mm}{r^2}$

I am using M for the mass of the large planet, and m for the mass of the projectile. Of course, Mm, otherwise we’d have to deal with the movement of both objects.

According to Newton’s second law, F=ma, which we can rearrange to a=F/m:

$\displaystyle a = \frac{F}{m}$

$\displaystyle a = -G\frac{Mm}{r^2}\frac{1}{m}$

$\displaystyle a = -\frac{GM}{r^2}$

As expected, the acceleration does not depend on the mass of the projectile. Let’s put this into our equation for the position.

$\displaystyle y = -\frac{1}{2}\frac{GM}{r^2}t^2 + r$

$\displaystyle y = -\frac{GMt^2}{2r^2} + r \ \blacksquare$

# Step 2: Find the x position on the circle corresponding to the y position

By the Pythagorean theorem,

$\displaystyle x^2 + y^2 = r^2$

$\displaystyle x^2 = r^2 - y^2$

$\displaystyle x = \sqrt{r^2 - y^2}, \ x\geq 0$

I’m going to assume the clockwise direction for orbit and will not consider the x<0 solutions. Let’s substitute in our expression for y from Step 1.

$\displaystyle x = \sqrt{r^2 - \left({-\frac{GMt^2}{2r^2} + r}\right)^2}$

This is rather complex, and I’m not sure if I should multiply this out or if something will otherwise cancel.

$\displaystyle x = \sqrt{r^2 - \left({\frac{G^2M^2t^4}{4r^4}-2\frac{GMt^2}{2r^2}r + r^2}\right)}$

$\displaystyle x = \sqrt{r^2 - \left({\frac{G^2M^2t^4}{4r^4}-\frac{GMt^2}{r^2}r + r^2}\right)}$

$\displaystyle x = \sqrt{r^2 - \left({\frac{G^2M^2t^4}{4r^4}-\frac{GMt^2}{r} + r^2}\right)}$

$\displaystyle x = \sqrt{r^2 - \frac{G^2M^2t^4}{4r^4}+\frac{GMt^2}{r}- r^2}$

$\displaystyle x = \sqrt{r^2 - r^2 + \frac{GMt^2}{r} - \frac{G^2M^2t^4}{4r^4}}$

$\displaystyle x = \sqrt{\frac{GMt^2}{r} - \frac{G^2M^2t^4}{4r^4}}$

That is an unfortunately complicated expression. I don’t see how I can simplify this, aside from factoring out the first term to get

$\displaystyle x = \sqrt{\frac{GMt^2}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)}. \ \blacksquare$

# Step 3: Find the horizontal velocity needed to get to the designated x position in time t

The position of x is given simply by

$\displaystyle x = \frac{1}{2}a_xt^2 + v_{0,x}t + x_0$

We’re assuming over this short period that the acceleration is downward, that there is no acceleration in the x direction.

$\displaystyle x = v_{0,x}t + x_0$

The initial velocity is entirely in the x direction (recall that the y component was 0 above), so we can write v0,x=v0

$\displaystyle x = v_0t + x_0$

And the projectile starts from a position of x = 0:

$\displaystyle x = v_0 t$

This is the basic distance-velocity-time equation, and is the usual partner to y projectile motion equation above. We solve for velocity:

$\displaystyle v_0 = \frac{x}{t}$

and plug in our expression for x from Step 2 above:

$\displaystyle v_0 = \frac{\sqrt{\frac{GMt^2}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)}}{t}$

Note that √(t²) = |t|, so if we assume t is positive, we can bring it into the radical without worrying about a plus/minus sign. I am interested in the time moving forward, not the time in the past when it could have gone from this position up to be at the “apex” at time=0.

$\displaystyle v_0 = \frac{\sqrt{\frac{GMt^2}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)}}{\sqrt{t^2}}, \ t\geq 0$

$\displaystyle v_0 = \sqrt{\frac{\frac{GMt^2}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)}{t^2}}$

$\displaystyle v_0 = \sqrt{\frac{GMt^2}{rt^2} \left({1 - \frac{GMt^2}{4r^3}}\right)}$

$\displaystyle v_0 = \sqrt{\frac{GM}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)} \ \blacksquare$

I don’t think I can further simplify this. Time for Step 4.

# Step 4: Evaluate for a very short time interval

Feynman used one second for his rough calculation. I could do the same, but I’m favoring a more mathematically rigorous approach. Why stop at one second? We could evaluate at one millisecond, or one nanosecond. I’ll try taking the limit as t goes to 0. I hope this results in a viable expression, and doesn’t just indicate that in an infinitesimal time, the projectile will hardly drop so any velocity works or something like that.

$\displaystyle v = \lim_{t \to 0} v_0$

We can plug in our expression for v0 from Step 3.

$\displaystyle v = \lim_{t \to 0} \sqrt{\frac{GM}{r} \left({1 - \frac{GMt^2}{4r^3}}\right)}$

$\displaystyle v = \sqrt{\frac{GM}{r} \left({1 - \frac{GM\cdot 0^2}{4r^3}}\right)}$

$\displaystyle v = \sqrt{\frac{GM}{r} (1)}$

$\displaystyle v = \sqrt{\frac{GM}{r}} \ \blacksquare$

This is a wonderful result: it looks just like the type of answer I was hoping for, with an appropriate amount of complexity (it “feels right”). Of course, there is probably a more straightforward way to derive it. Let’s test it a real-world example, the orbit of the International Space Station.

The ISS doesn’t have a perfectly circular orbit. According to Wikipedia, the perigee and apogee (minimum and maximum distances) are 409 and 416 km, respectively, so I’m just going to take 412.5 km as an average distance (I don’t really know what the appropriate “average” distance for an elliptical orbit would be). This altitude will be on top of the radius of Earth, about 6371 km. Plugging these values in, along with the mass of Earth and the gravitational constant, gives

$\displaystyle v = \sqrt{\frac{(6.674\times 10^{-11} \ \mathrm{N\cdot m^2\ kg^{-2}})(5.972\times 10^{24}\ \mathrm{kg})}{(6371 \times 10^3 \ \mathrm{m} + 412.5 \times 10^3 \ \mathrm{m})}}$

$\displaystyle v = \sqrt{\frac{(6.674\times 10^{-11} \ \mathrm{N\cdot m^2\ kg^{-2}})(5.972\times 10^{24}\ \mathrm{kg})}{(6783.5 \times 10^3 \ \mathrm{m} )}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\cdot \frac{10^{-11}\cdot 10^{24}}{10^3}\cdot \frac{\mathrm{N}\cdot\mathrm{m^2}\cdot\mathrm{kg}}{\mathrm{kg^2}\cdot\mathrm{m}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5} \cdot \frac{10^{-11}\cdot 10^{24}}{10^3}\cdot \frac{\mathrm{N}\cdot\mathrm{m^2}\cdot\mathrm{kg}}{\mathrm{kg^2}\cdot\mathrm{m}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\times 10^{10} \cdot \frac{\mathrm{N}\cdot\mathrm{m^2}\cdot\mathrm{kg}}{\mathrm{kg^2}\cdot\mathrm{m}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\times 10^{10} \cdot \frac{\mathrm{N}\cdot\mathrm{m}}{\mathrm{kg}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\times 10^{10}\cdot \frac{\mathrm{\frac{kg \cdot m}{s^2}}\cdot\mathrm{m}}{\mathrm{kg}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\times 10^{10} \cdot \mathrm{\frac{kg \cdot m^2}{s^2 \cdot kg}}}$

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}\times 10^{10} \cdot \mathrm{\frac{m^2}{s^2}}}$

I am relieved that the units worked out.

$\displaystyle v = \sqrt{\frac{6.674 \cdot 5.972}{6783.5}}\cdot\sqrt{10^{10}} \cdot \sqrt{\mathrm{\frac{m^2}{s^2}}}$

$\displaystyle v = 7.67\times 10^{-2}\cdot 10^{5} \cdot \mathrm{\frac{m}{s}}$

$\displaystyle v = 7.67\times 10^{3} \ \mathrm{m/s} \ \blacksquare$

According to Wikipedia, the ISS orbits at an average velocity of 7.66 km/s, or 7.66×10³ m/s. Our approach worked! To be honest, I wasn’t expecting this degree of precision. This is fantastic!