Kerbol Gravity Question

[TheAngryHulk]

I just started the game a couple weeks ago and am just now getting used to all the math (don't have mods yet). My question is, in all the videos and discussions I've found I see people using 9.81 m/s2 in their math. Why is that number used when the wiki says Kerbol has a gravity of 17.1m/s2?

 

EDIT: NVM, I just realized Kerbol is the sun not the planet. Sorry. I tried to delete this but couldn't find a way

Edited May 23, 2018 by TheAngryHulk

[Dman979]

Well, I'll answer this question anyway.

I assume you're talking about the formula for Delta-V (=I_sp*g₀*ln(m₀/m_f)). Let's break that down into it's component parts.

Delta-V is the potential speed change your rocket can produce. Since most of the time a rocket is coasting, you only need to change your speed when you want to coast to a different place.

I_sp(also commonly seen as Isp, isp, ISP, or anything similar) is the efficiency of your rocket- how many miles-per-gallon it would get, if it were a car.

We'll skip g₀for now, and come back to it later.

ln(m₀/m_f) measures what percentage of your rocket can be burned by the engines, and then takes the natural logarithm of that.

m₀is the total mass of your rocket at Time Zero- the current mass of the rocket, including all the fuel, tanks, piping, wires, engines, etc.

m_f is the final mass of your rocket. This is the same as m₀, except there's no fuel left- only the tanks, piping, wires, engines, etc.

 

Now, let's come back to g₀. Normally, when you see a letter in an algebra equation, it stands for a variable- a number whose value is subject to change, depending on the conditions. This equation, though, is a physics equation. Why does that matter? Well, physics has some slightly different rules about letters in equations. In this case, g₀is a constant- a number whose value does not change.

And g₀ isn't just any constant- it's the standard acceleration due to gravity at the Earth's surface. It doesn't change if you're at a different planet, since it's defined at Earth's surface: it will always be the same.

So, what is the standard acceleration due to gravity at the Earth's surface? 9.80665 m/s² -- also known as 9.81 m/s².  

Hope this helped.

Also, I'm moving this to Gameplay Questions.

[OhioBob]

g_o is used in the equation because that's the way specific impulse is defined.  Specific impulse is the ratio of thrust to the flow rate of weight ejected.  In imperial units, this is pounds thrust divided by pounds per second of exhaust product.  (The pounds cancel so we end up with specific impulse having units of seconds.)  But what is weight?  Weight is mass times the acceleration of gravity.  But we want specific impulse to be a standard unit that is the same everywhere.  We don't want it to vary depending on where we are, so we define the weight to be that on the surface of Earth.  Therefore we multiply the mass flow rate by earth's standard gravity to get weight.  The equation for specific impulse is, therefore

I_sp = F / (ṁ * g_o)

where I_sp is specific impulse, F is thurst, ṁ is the mass flow rate, and g_o is standard gravity (9.80665 m/s²).

The rocket equation is,

Δv = V_e * LN( m_o / m_f )

where V_e is the effective exhaust gas velocity.

Thrust is the effective exhaust gas velocity times the mass flow rate,

F = V_e * ṁ

So if we substitute V_e * ṁ for F in the specific impulse equation we get,

I_sp = (V_e * ṁ) / (ṁ * g_o)

Canceling out ṁ and rearranging, we get

V_e = I_sp * g_o

So that's how we end up using specific impulse times standard gravity in the rocket equation.

 

Edited May 24, 2018 by OhioBob

[Fraston]

Hey, welcome to the forums!

[TheAngryHulk]

Thank you all for the replies. I actually forgot I posted this until today. Sorry for the late reply.