acceleration due to gravity and velocity of a crash

[ummwut]

I want to calculate the velocity of impact upon a surface of a body from some altitude, and an initial downwards velocity. Maybe also a time to impact, as well.

Normally I wouldn't need to ask for help with something like this, but I don't know enough physics to get through this. I know the acceleration is changing with some change in distance, which can be integrated into a total acceleration over a distance, but from there I am stuck.

This is assuming a heading of straight down, no atmosphere, and the body is not rotating underneath.

Edited April 30, 2013 by ummwut

[Warscribe]

from what i can tell, that is very dependent on your trajectory and how high up you are (so the rotation of the celestial body you're aiming at comes into play, also the gravity acceleration). Then it depends on the thickness and makeup of the athmosphere (reentry burns and air resistance), how big and heavy your craft is, how the weather on said celestial body is and how high from sea level your impact surface is.

thats a whole lot of variables.

[ummwut]

I'll edit my first post to give details, but this is assuming no atmosphere, and heading straight down to the center of the body with no rotation happening.

I need to go learn about physics.

[falofonos]

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

[SunJumper]

For short falls, just use v^2 = u^2 + 2as.

[Warscribe]

well if you said youd integrate to overall acceleration than it would just be v=t*a for traveling velocity, where t is time and a is acceleration. Reverse engineer this. distance/t = t * a ; solve for t. (i'm lazy.); t = +/- (sqrt distance) / (sqrt a) tahts your impact time from the moment you accelerate towards point of impact.

man i havent done this in a long time...so i may as well be wrong.

Edited April 30, 2013 by Warscribe

[ummwut]

The game is huge, there is no such thing as a "short fall".

Thanks falofonos, I didn't think to consider a falling ship to be a falling projectile!

Oh, but that assumes constant acceleration :/

Edited April 30, 2013 by ummwut

[SunJumper]

The game is huge, there is no such thing as a "short fall".

Thanks falofonos, I didn't think to consider a falling ship to be a falling projectile!

Oh, but that assumes constant acceleration :/

Falling 5km after killing horizontal velocity on the Mun is classed as a short fall in this case. It's just when the change in acceleration due to gravity is negligible...

[ummwut]

SunJumper, where did you get this formula? Is there a relevant wikipedia page?

[falofonos]

Roughly how large a fall are you looking to calculate?

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Gravitational_field

equations can be used to determine gravitational acceleration

If your fall is from outside the sphere of influence of the target body then I'm not even sure the game simulates that gravitational force, also the maths get horrible... probably why the game doesn't simulate it.

May be easier to use energy conservation.

Potential energy of your rock = (-(gravitational constant)x(mass of target planet or moon)x(mass of rock)) / (distance of separation between the two)

run 2 calculations one for starting position and second for impact using the planet's radius as separation treating both objects as point masses just for simplicity.

units are meters and kilograms.

at impact the kinetic energy of your projectile will be equal to the difference between the 2 potentials.

KE= (mass x velocity^2) / 2

so your velocity will be the square root of (2x potential difference) / mass ... I think.

Edited April 30, 2013 by falofonos
been a while since I did any physics....

[Warscribe]

you just need the "average" acceleration, as integrated: int from 0 (sea level) to d(total distance from sea level) with the formula on how the gravitational acceleration changes (its probably something logarithmic, like ln(x))

[ummwut]

@falafonos:

The fall is from within the SOI, no worries there. But it is definitely on the order of hundreds of thousands of kilometers.

@Warscribe:

The total acceleration is some constant times (1/low altitude - 1/high altitude). I believe the constant is whatever the gravitational acceleration is at the center of the mass.

[trawum]

Assuming there's no air resistance, all you'd need is the specific energy. KE = 1/2v^2 and PE = -(gravitational parameter)/(altitude) at any point. So you'd get something like this

Where mu = gravitational parameter, R = planetary radius, A = initial altitude, and v = initial speed.

[The Lone Wolfling]

If you're in an atmosphere, most of the time it doesn't really matter how far you fall, you'll hit going at about (rather, slightly above) terminal velocity.

For planets without atmosphere, see trawum's post, above.

[ummwut]

Assuming there's no air resistance, all you'd need is the specific energy. KE = 1/2v^2 and PE = -(gravitational parameter)/(altitude) at any point. So you'd get something like this

Where mu = gravitational parameter, R = planetary radius, A = initial altitude, and v = initial speed.

How did you derive this?

[NikkyD]

sorry for bringing this old thread up again but...

has this been answered anywhere to anyones satisfaction ?

What i'm trying to calculate is, when i enter muns SOI with speed v_0 going straight "downwards", with what speed will i hit the surface with (and thus, how much delta_v will it cost me to land)