# Satellite Motion conceptual problem

1. Jan 5, 2014

### godingly

Hey Dear forum,
I know I'm wrong, but I don't know why:

Since the Earth's descends 5 meters for every 8km, a horizontal speed of 8km/sec will put an object to rotate around the Earth (neglecting air resistance). I.e. in one second it will fall 5m below it's initial height, but so will the earth. But by the simple formula S = V$_{o}$t + $\frac{at^{2}}{2}$ by the 2$^{nd}$ second it suppose to descend 20 meters from its original height - but the Earth would only curve by 10 (5*2)!! Which means during the 1st and 2nd second it will crash into Earth!

What is my mistake?

2. Jan 5, 2014

### voko

Your mistake is in your assumption that the field of gravity has a constant direction while the object travels. That is approximately correct for short distances, but you are trying to use the approximation for larger distances.

A better approach is through asking the following question: what velocity must an object have to rotate circularly with radius equal to that of the Earth and centripetal acceleration equal to the acceleration of gravity at the surface of the Earth.

3. Jan 5, 2014

### AlephZero

That. Relative to your starting position, the earth curves approximately 5 * 22 m in 16 km. Draw a diagram of the arc of a circle and a line tangent to it.

Or, as Volo said, if you want to consider each interval separately, you have to remember that "downwards" always points towards the center of the earth, so it is in a different direction in the two intervals.

4. Jan 5, 2014

### sophiecentaur

You appear to be trying to impose 'flat Earth' calculations on a spherical earth, here. The parabola equation only works properly for a uniform g field over a horizontal surface. I could take you a little way over the horizon before it fails but it definitely isn't the way to approach Orbital mechanics.

To work out that, you need to look into the force need to produce circular motion. That force is a 'central attractor' and is provided by gravity, which follows the inverse square law. The 'book work' that shows you what's what can be found in many places. To maintain a particular altitude of orbit, you need to be travelling at a certain speed, which will require a particular centripetal force (equal to the g force at that height).
The guys on Hyperphysics are pretty good as a source of this sort of basic theory.

5. Jan 5, 2014

### Staff: Mentor

The parabola is fine to consider the first few seconds - the important point is the quadratic deviation from a flat surface. 5m in 8km imply 20m in 16km, not 10.

6. Jan 5, 2014

### sophiecentaur

Maybe but when the title of the thread includes the word 'satellite', a flat earth / parabolic approximation is hardly the right thing.
The introductory animation, showing a gun firing a shell horizontally with progressively higher and higher velocity until an orbit is achieved, is fine but not if anyone wants to assume parabolic motion.

With or without an arithmetical mistake, the quantitative approach is a bit too flawed to make it worth while, imo. Arm waving about it is fine.

7. Jan 5, 2014

### godingly

Thank you all for the answers! For mfb and alephzero - is there a formula to calculate the perpendicular (normal) distance from a point on a tangent line to the circle? When I say normal, I mean 90 degrees at the tangent line, not at the circle.

8. Jan 5, 2014

9. Jan 5, 2014

### voko

The formula is $$h = r \left(\sqrt {1 + \tan^2 \frac s r} - 1\right),$$ where $r$ is the radius, and $s$ is the arc distance.

10. Jan 5, 2014

### AlephZero

Or $$h = r\left(\frac{1}{\cos{\dfrac s r}} - 1\right)$$ which is fewer buttons to push on most calculators.

Note, in this and Volo's formula, the trig functions use radians not degrees.

Or if $s$ is small relative to $r$, approximately $$\frac{s^2}{2r}$$

11. Jan 5, 2014

### voko

I missed an obvious simplification, dammit!

12. Jan 6, 2014

### sophiecentaur

The shape of the curve that's followed is an ellipse - just as with all orbits. The ellipse will pass through the Earth if the velocity isn't right. Needless to say, if you take off from a point on the Earth's surface (the highest mountain you could find), the 'successful' orbit (neglecting gravity), will actually be a circle (special case of an ellipse) and will take the shell round and it will hit you in the back of the head after about 90 minutes.