Determining Launch Velocity to Overcome Drag (no propulsion)

[TimNJ]

Hi everyone,

I'm working on a project for my electromagnetics class which asks us to design an electrostatic launch system that can propel an object into space. Though the bulk of the project is supposed to be about designing the accelerator plate array, I got really interested in the whole concept along the way.

Homework Statement

We have to propel a 0.1kg golfball sized object into space. Since air resistance at low altitudes is very high, I decided to launch my object from an altitude of 35km using a high-altitude (weather) balloon.

2. The attempt at solving the problem

In Excel, I did some numerical analysis which outputs the altitude of the ball after launch with respect to time. This analysis takes both drag and gravity into consideration.

However, I found that a wide range of launch velocities will get the ball past 100km (the Karman line), but I also know that the object must be launched at a velocity at least the “escape velocity”. On the surface of the Earth, this velocity is ~11.2km/s, and at my altitude it only drops to 11.15km/s. So, at the very least my launch velocity must be 11.15km/s or else gravity will pull the object back to earth.

My dilemma is finding what the launch velocity should be accounting for wind resistance (drag). At 35km, rho, the density of air is about 100 times less than near the Earth’s surface. It’s about 0.011kg/m3, but its still not negligible if you launch at >11.15km/s.

I’ve seen this approach: ΔVlaunch = ΔVgravity + ΔVdrag. I’m pretty sure I have ΔVgravity which is 11.15km/s, but I’m not sure how to find ΔVdrag.

3.) Related Questions

How can I determine the additional launch velocity I will need to overcome drag, given the launch altitude etc.? Would it be a numerical answer? Perhaps some sort of integration is involved?

Thanks so much!

 

[Simon Bridge]

Presumably the range of launch velocities is everything from some minimum to the speed of light?
Escape velocity, without other losses, will get you from the surface of the Earth to well into the solar system - probably ending up in the Sun.
To get from 25km up to 100km up you need the kinetic energy equal to the potential difference plus whatever air resistance will remove.

Working out drag:
http://www.engineeringtoolbox.com/drag-coefficient-d_627.html
... you work out the drag force - and include that in your numerical computation.

 

[TimNJ]

Thanks,

So I have KE = PE(gravity) + W(drag). That is, 1/2mV² = GMm/r + W(drag).

Now W(drag) is the integral of F(drag), which should equal F(drag) * distance, which is 0.276v², in my case. I got that by multiplying Δx, (100km-35km=65km), by the drag coefficient I calculated at 35km, 4.257 x 10^-6.

However, work done by drag is still a function of velocity squared, so how might I approach this problem? Do add up all the bits of work done between 35km and 100km? Also, at 100km Earth's gravity is still in effect. (Hell at 100,000km it's technically still in effect). So am I wrong to even factor 100km into the equation at all? Thanks.

 

[Simon Bridge]

Your model for drag assumes a constant density for the air, this is not the case over the distance involved ... and yes, you add up all the bits of drag at each step, which will involve an integral.
That or you do it numerically.

Please follow the link.

 

[TimNJ]

Thank you Simon,

So if drag is dependent on the velocity squared, but I don't know what my velocity will be vs time and at different altitudes, how can I compute this? That is, how can I figure out what my ΔV(drag) is when drag itself is a function of velocity?

Am I missing something in the link? All I can find from it is that my ball will have a drag coefficient of ~0.5 and that W = Fd.

Thanks.

 

[gneill]

Looks like you'll be needing to use numerical integration (simulation) to determine the trajectory. If you just want to find the launch speed that gets you to a certain altitude or escape from the Earth then you can create a numerical model and use trial and error to locate the threshold launch speed that gets you the desired result. This involves taking small time steps or distance steps, assuming that certain parameters remain essentially constant over those small steps, and summing the effects of the individual steps to create an overall trajectory.

If you just need a pass/fail sort of test for a given initial launch speed then you might consider using small distance steps and working with energy changes (ΔKE, ΔPE, ΔE due to friction (drag)). If there's KE remaining at your target height then your initial speed passes the test.

At a given instant a projectile has a position (radial distance from the center of the Earth), a kinetic energy KE (velocity) and a gravitational potential energy PE. There is also the local atmospheric density ρ at that elevation with which, along with the instantaneous velocity, you can find the current drag force. If you examine a small change in distance Δr then you can use energy considerations to determine the KE at the new position Δr further along the trajectory. If the distance step is small you can assume that the drag force is constant over the interval (or you can get fancy and use an average over the interval, since you can work out the density at both ends or the middle of the distance step). The energy lost to friction follows from the drag force and step size. If the KE goes to zero or goes negative after the step, you've reached or passed the maximum attainable distance for you initial launch speed.

The NASA website has a simple atmospheric model that can give you the density at a given altitude. Unfortunately it uses English units, so you'll have to do some unit conversions (don't crash your lander ).

Good Luck!

 

[Simon Bridge]

Yep - you need to use a numerical method.
The link gives you a formula for drag force - you also know the formula for gravity force and Newton's laws.

If y is altitude, then $$\frac{dW}{dt} = \frac{1}{2}C_dA\rho(y)\left(\frac{dy}{dt}\right)^3$$ ... where W is the energy lost to drag.
I think (used dW = Fdy and divided through by dt).
You could also go: $$\frac{1}{2}C_dA\rho(y)\left(\frac{dy}{dt}\right)^2 = m\left(g-\frac{d^2y}{dt^2}\right)$$ ... from Newton's second law, taking "up" as positive.

You will need to google the terms, like how the air density varies with altitude.

You know the expression "it ain't rocket science"? ... well this is rocket science. Enjoy.