Michael D. answered • 06/13/19
Such a fun question...
The first component is to calculate escape velocity or in fact define escape speed. We need to define escape speed as the speed required to attain at infinite distance from the potential well a final kinetic energy and potential energy for our rocket to be zero. Using conservation of energy- the kinetic and potential energy of the rocket at this speed must equal the kinetic and potential energy of the rocket at this infinite distance.
The potential energy of an object in a gravitational well of mass M and at distance r is -GM/r. So at infinite distance this is zero and at the surface of the earth this is -gre. Where the acceleration at the surface of the earth is GMe/re2 which is 9.8 m/sec-sec. Then conservation of energy for an escape speed at the surface of the earth, neglecting drag is ve=SQRT(2gre) from Kinetic energy = Potential energy when this speed is attained, and we are assuming that our rocket attains this at the surface of the earth.
The diameter of the earth at the equator is 12,756 Km yielding an escape speed of 11.181 Km/sec.
Now part b either you have been given the Tsiolkovsky rocket equation or tools to derive it using some simple calculus concepts ?
Newton's second law sum of forces equals dP/dt where P is the total momentum vector of our rocket system.
The sum of forces on our rocket is zero and leads to to a differential equation:
dV/dt =-ve (1/m) dm/dt this expression on integration of our rocket velocity and loss of mass due to propellant exhaust leads to ΔV = veln(mo/mt) mo being the rocket with initial fuel mass and m(t) being the rocket mass after the fuel has been expended to reach the escape velocity.ve is the propellant velocity
One can define Mf as the propellant mass fraction of our rocket...( mo-mt)/ m0 and write this as the expponential decay ( 1- eΔV/ve) Then this mass fraction multiplied by the initial mass of the rocket is the mass of fuel expended on reaching the escape velocity.
For example lets say our rocket has a propellant velocity equal to the escape velocity. The mass fraction is 1-e-1 = 0.632 and the mass of fuel for our rocket having total mass of 450 kg is 284 kg. The initial velocity of our rocket is zero and its final velocity is the escape velocity.
part c investigates impulse or specific impulse for our rocket...the time required to exert an average force that yields this change in momentum. The exhaust velocity is the acceleration due to gravity multiplied by the specific impulse in seconds.
Faverage = dP/dt delta P is what we call impulse, F average is mg for earth. mgt = impulse.
Impluse/mass is specific impulse = gt
If the exhaust velocity is the escape velocity we write the specific impulse as 11,181 m/sec/ 9.8 m/sec-sec or lets use 10 m/sec-sec for g....1,181 sec for this impulse in seconds.
The mass rate of loss...assuming uniform loss is 284,000 g/1,181 sec or 240 g/sec
In part b and c I have not provided the result for the exhaust velocity of 4250 m/sec
It seems like the writer of this problem has tried to obfuscate the solution by including the mass of earth.
The mass provided is correct but only has bearing in the calculation of g which we should already know to be 9.8 m/sec-sec. It would seem more useful in the calculation of the escape velocity to have been given the diameter of the earth. Although, I suppose with the mass and g = GM/r2 that can be determined. The universal gravitational constant is listed.