Gravity doesnt match with real physics? Or am I confused xD?

[LokoGz]

Wondering (as im studying it...) what was Gilly gravity acceleration, and doing math with this: "y=y0+v0t+1/2at^2", with the info in the video; (y=0, y0=150m, t= ~100s) i get around 2.8m/s^2 and the wiki says 0,04m/s^2; so i ask:

Is there other ecuation to know the gravity (surely) or this ecuation just doesnt work with this kind of things or... i dont know. Am i confused xDD?

Edited October 9, 2014 by LokoGz
removed a tempting typo

[Hannu]

You made some error in calculation.

s=0.5*a*t^2

a=2*s/t^2 = 2*150 m /(100 s)^2 = 0.03 m/s^2.

[KerTrotsky]

Should have fired the rocket pack for vertical accel. Would have been a better test.

[KerikBalm]

v_o = o, so its just x=1/2 a t^2

x=150, t = 100

150 = 1/2 a * 10,000

300 = 10,000a

300/10000 = a

3/100 = a

a= 0.03

However, you also have to keep in mind that a is not constant.

On Earth, with a radius of over 6,000 km, 150 meters is not a big change in distance from the center of mass

On Gilly, with a radius of 13km, being on a mountain 2km high will have a significant effect on the surface gravity, and being .15km above that will also affect the gravity

its f= G * (M_1) * (M_2)/r^2

This treates the masses as point masses, and since f=ma, we can get acceleration of M_2 and drop it out of the equation

anyway since KSP also uses the simplified assumption of a point mass at the center of gravity (instead of "lumpy" gravity fields that vary with the amount of mass below them... like when a spacecraft passes over a mountain for instance),

If you were at elevation = 0 on Gilly, and then jumped 150m high, the ratio of the strength of gravity at the top vs bottom would be (13,000)^2/(13,150)^2 = 0.9773

ie, it would be just over 2% weaker.

But climb up to 2800 m, as you do in the video, it is now 13000^2/(13000+2800)^2= 0.677 ... about 2/3 of the strength at elevation=0

.049*0.677 = 0.0332.. ie basically 0.03

But those numbers we used were rounded - the actual change in height was 159 meters, not 150, which will increase the calculated a, which is good because the calculated a was a little low.

anyway, the wiki seems fairly accurate

Edited October 3, 2014 by KerikBalm

[cantab]

The Gravioli detector can give you a readout of current acceleration. Use it when landed on Gilly and there's your measurement.

[Claw]

Thanks for the point out about the typo guys, but there's no need to pull the thread off topic to a prohibited topic.

Thanks,

~Claw

[LokoGz]

Ok, so t^2 is where I failed, ty all.

[capi3101]

You can also find the gravitational acceleration at any given altitude above the body with this equation:

g = GM/(R^2), where g is the gravitational acceleration, G is the gravitational constant, M is the mass of the body, and R is the distance to the center of the body's mass.

GM is sometimes represented in aggregate by the greek letter Mu, and is a parameter called the standard gravitational parameter. For the bodies in the Kerbol system, all of this information is readily available from the wiki. So, for Gilly at its surface -

mu = GM = 8,289,449.8 m^3/s^2

R = 13000 m (i.e. the moon's equatorial radius)

g = (8,289,449.8 m^3/s^2) / (13000 m)^2 = (8,289,449.8 / 169,000,000) * (m^3/m^2s^2) = 0.049049999 m/s^2

Bear in mind that this would be the value at "sea level" (i.e. an altimeter of zero meters). For 2800 meters above sea level, the value becomes:

g = (8,289,449.8 m^3/s^2) / (15,800 m)^2 = 0.033207 m/s^2 (i.e. what everybody else is telling you).

EDIT: And I see that KerikBalm told you this a few posts ago. So never mind me...