## Saturday, May 17, 2014

### Geostationary

Satellites in a geostationary orbit appear to maintain a fixed position relative to a position on the surface of the Earth. To accomplish this, they need to rotate at the same angular speed that the Earth spins, which makes one full rotation every sidereal day. There are 23.93446122 hours (or 86,164.0604 minutes) in one sidereal day. A full revolution of 2π radians in 86,164.0604 seconds gives us $$\omega = 7.29212 \times 10^{-5} \ radians \cdot s^{-1}$$. This is our target angular speed if we want to keep a satellite overhead the same spot.

Earth's standard gravitational parameter is its mass M x G (the gravitational constant), which is roughly μ = 398,600.4418. The standard gravitational parameters for planets are better known than either M or G individually (the best way we have at our disposal to weigh extremely massive objects is by observing their gravitational effect on other bodies like orbiting satellites). Acceleration due to gravity is $$\mu \over r^2$$. The equatorial radius of Earth is 6,378.14 km. If we plug in that figure, we'd see acceleration on the surface is the familiar 0.009798285 km/s. 422 km above the surface (in the orbital sphere of the ISS), it's just 0.008619904 km/s (87.97%). The further out we go, the lower the acceleration and the longer the orbital period.

To find out how far away we have to place a geostationary satellite, consider that it will need to be in a circular orbit with an orbital period of one sidereal day. In one second, it will move $$7.29212 \times 10^{-5} radians$$; therefore $$r \times (1 - cos(7.29212 \times 10^{-5}))$$ will be the distance that it falls towards the Earth in that time. We know from the SUVAT equation $$s = ut + {1 \over 2}at^2$$ that in a second, with no initial velocity, that the distance travelled is half the acceleration. We now have an identity $$r (1 - \cos(7.29212 \times 10^{-5})) = {\mu \over r^2}$$. Rearrange it and solve: $r = \sqrt[3]{\mu \over {2 - 2 cos(7.29212 \times 10^{-5})}}$ It turns out that the altitude of 42164.15974 km above the center of the Earth (35786.02274 km above the equatorial surface) is where you'll find geostationary satellites - and they can only exist in that one orbital plane.

## Saturday, May 10, 2014

### Gravity / ISS

Let's say we're interested in finding how fast the International Space Station has to move in order not to fall back to earth (i.e. to stay in orbit). The earth has an equatorial radius of about 6384km and the satellite is about 422km above the equator when it's overhead on its inclined orbital path. For the sake of simplicity we will ignore the oblate spheroid shape of the earth, and the replace the elliptical orbit with a circular one, 6806km from the center of the earth.

On Earth's surface the acceleration due to gravity is $$9.8 m \cdot s^{-2}$$. The further you go away, the lower the force of gravity. 422km above the surface we're told that the acceleration is 89% of surface gravity - it's $$8.722 m \cdot s^{-2}$$.

Using the SUVAT equation $$s = ut + {1 \over 2}at^2$$ we can see that an object dropped from 422km (i.e. $$g = 8.772 m \cdot s^{-2}$$) would fall 4.361m in the first second.

In the same second, we know that the ISS traces out the circular orbit (i.e. it doesn't crash into the Earth). The angular distance it travels (we could use the unit circle for visual confirmation) is $$\arccos({{r - s} \over r})$$ or $$\arccos({{6806000 - 4.361} \over 6806000})$$ or $$\arccos({6805995.639 \over 6806000})$$ or simply 0.001132041 radians.

If it travels 0.001132041 radians in a second, it will take 5550.316851 seconds to perform a complete revolution of 2π. 5550 seconds is 92½ minutes, a figure that's very close to the published figure (on Wikipedia) and that's quite amazing given the rough estimates we've made to get this far. A circle of radius 6806km has a circumference of 13612km. Divide that circumference by the orbital period in seconds and you get $$7704 m \cdot s^{-1}$$. You could also try $$r \sin(\theta)$$ or $$6806000 \sin(0.001132041)$$ which gives the same result. It's moving pretty quickly, but would have to be even quicker if the orbit was any closer to the surface!

## Wednesday, April 16, 2014

### Newtonian Reflector

Newtonian reflector telescopes use two mirrors – a figured primary mirror (the objective) which focuses incoming light, and a flat secondary mirror which redirects that light at an angle so that it can be viewed without getting your head stuck in the tube. The secondary mirror is only practically important – a CCD sensor could be placed directly into the path of the light focused by the objective mirror and it would capture the light without requiring an additional reflection. However, the subject of this post is the primary mirror, particularly: it’s shape.

A 2D parabola is defined by the quadratic equation $$y = ax^2 + bx + c$$. If we make some simplifying assumptions (such as: its open side faces up; it “rests” on the x-axis; the focus is on the y-axis at height p) then our equation reduces to $$y = 4px^2$$. The first order derivative (e.g. slope) of this equation is $${dy \over dx} = 2px$$. That means, for a given focal length (and radius from the centre of the mirror), we can estimate the height of the reflective surface from the x-axis, as well as the slope at that point. These two computed values show where light is reflected when it encounters the surface of the objective mirror. Remember that incident light is reflected about the normal vector to the surface:
$$R = D – (2D \cdot N)N$$

To enable a 3D version (a paraboloid) we can take advantage of the knowledge that we only need rotate the parabola (i.e. the shape is rotationally symmetric about the vertical axis). In 3D, I chose to rotate around the z-axis.

It then becomes rather straight forward to take a solid angle of light, to compute the angles at which each incident ray encounters the parabolic reflector surface (parallax will affect closer light sources more than further away sources) and then note the points at which two or more rays converge to a single point. It turns out that the parabolic shape is ONLY able to focus light that is travelling parallel to the reflector’s primary axis. Closer sources converge further back than the stated focal length; sources at infinity converge exactly at the parabola’s focal point. Off-axis sources result in a complex out-of-focus shape when we attempt to capture them on a flat focal plane like a CCD. Looking at the distribution of focal points for a given field of view it was difficult to imagine a single transformation that might enable coma to be completely (and correctly) removed, though clearly a retail market for coma-correcting lenses abounds.