# Optimal ground launch flight path

## Contents

## Introduction

The current derivation of the rocket equation on wikipedia (Aug 13 2009) seems to be overly cumbersome: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

The rocket equation basically comes down to the basic law of Newtonian dynamics. That is the sum of the forces equals the mass multiplied by the acceleration.

<math>\sum F=ma</math>

The product of the rocket exhaust velocity multiplied the mass flow is the rate momentum is leaving the rocket. Because momentum must be conserved the rocket gains an equal and opposite amount of momentum as the exhaust. A more convenient form of newtons law is:

<math>\sum F=ma=m\frac{dv}{dt}=\frac{dmv}{dt}=\frac{dp}{dt}</math>

That is the force is the rate momentum changes with time. Consequently, the force exerted on the rocket by the exhaust is equal to the velocity multiplied by the mass flow. Taking into gravity the fores acting in the direction of flight and dividing both sides by the mass we obtain:

<math>a=V_e\frac{dm}{dt}+g sin(\theta)</math>

## Optimal Flight Path

The optimal flight path occurs when the vertical component of the thrust cancels out gravity and the remaining part of the thrust is due to horizontal acceleration. This can be written as:

<math>g-\frac{v^2}{R}cos(\theta)-T sin (\theta)=0</math>

where: g is the gravitational acceleration R is the distance from the center of the earth T is the thrust per mass.

Using trig identities:

<math>g+/-\frac{v^2}{R}\sqrt{1-sin^2(\theta)}-T sin (\theta)=0</math>

Rearranging:

<math>+/-\frac{v^2}{R}\sqrt{1-sin^2(\theta)}=T sin (\theta)-g</math>

Squaring both sides:

<math>\frac{v^4}{R^2}(1-sin^2(\theta))=T^2 sin^2 (\theta)-2gTsin(\theta)+g^2</math>

Rearranging:

<math>(T^2+\frac{v^4}{R^2}) sin^2 (\theta)-2gTsin(\theta)+g^2-\frac{v^4}{R^2}=0</math>

This equation can be solved using the Quadratic Equation.

## A Visual Basic Function to Calculate the Optimal Thrust Angle

Bellow is a visual basic function used in RocketDesignSpreadsheet to find the optimal thrust angle.

Function thrustAngle(V, T, Optional h, Optional g) Re = 6378.137 * 1000 ' Radious of The Earth If IsMissing(h) Then h = 0 End If If IsMissing(g) Then Gc = 0.00000000006674 ' N(m/kg)^2Gravational Constant Mearth = 5.9736E+24 ' Mass of the earth g = Gc * Mearth / (h + Re) ^ 2 End If R = h + Re a = (T ^ 2 + V ^ 4 / R ^ 2) b = -2 * g * T c = g ^ 2 - V ^ 4 / R ^ 2 des = b ^ 2 - 4 * a * c If des < 0 Then thrustAngle = 0 Else r1 = (-b + Sqr(des)) / (2 * a) r2 = (-b - Sqr(des)) / (2 * a) theata1 = WorksheetFunction.Asin(r1) theata2 = WorksheetFunction.Asin(r2) resid1 = g - V ^ 2 / R * Cos(theata1) - T * Sin(theata1) resid2 = g - V ^ 2 / R * Cos(theata2) - T * Sin(theata2) If Abs(resid1) < Abs(resid2) Then thrustAngle = theata1 Else thrustAngle = theata2 End If End If If thrustAngle < 0 Then thrustAngle = 0 End If End Function

## Total Gravity Losses

Returning to the equation mentioned in the introduction:

<math>a=V_e\frac{dm}{dt}-g sin(\theta)</math>

If we integrate this equation over time, we get the change the change in velocity we achieve. The term g sin(\theta) represents the gravity losses, and if we integrate this term over the flight path, it will gives us the additional delta V we need to overcome because of gravity losses. This term does not need to be calculated for the whole flight path because these losses will taper off as the thrust angle approaches horizontal.