Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Aug 4th 2015, 07:28 AM

#1  Junior Member
Join Date: Aug 2015
Posts: 2
 Calculating net thrust of rocket with gravity Trying to calculate the net thrust of a rocket taking into account the force due to gravity. So I presume this is the calculated Thrust minus the force due to gravity. Known data so far is that the rocket gets to escape velocity (11,200 m/s) after 5 minutes (300 secs). Acceleration for this is 11,200 / 300 = 37.3 m/s/s. Next needed to calculate the mass loss rate of fuel. Equation given for this was was q = ma / v where q is the mass loss rate, m = final mass of rocket, a = acceleration and v = fuel exit. m was given as 50,000 kg + 2000 Kg = 52,000 kg (without fuel) v was calculated from equation earlier with sqrt((2 * (P2  P1))/p), with given pressures and density. This came out as 2500 m/s So... q = (52,000 * 37) / 2500 = 770 kg/s From this the Thrust (T) = qv = (770 * 2500) = 1,925,000 N (It was assumed the pressures were the same so T = qv) Calculating the total mass of the rocket on launchpad needs the weight of the fuel which I presume is the mass loss rate (770 Kg/s) multiplied by the duration of fuel (300 secs) = 231,000 Kg Therefore the force due to gravity is... F = ma = (231,000 + 52,000) * 9.8 = 2,773,400 N. This is more than the thrust calculated earlier which would give a negative net thrust which cannot be correct. Can anyone see where I've gone wrong please? Regards
Trying to calculate the net thrust of a rocket taking into account the force due to gravity. So I presume this is the calculated Thrust minus the force due to gravity.
Known data so far is that the rocket gets to escape velocity (11,200 m/s) after 5 minutes (300 secs). Acceleration for this is 11,200 / 300 = 37.3 m/s/s.
Next needed to calculate the mass loss rate of fuel. Equation given for this was was q = ma / v where q is the mass loss rate, m = final mass of rocket, a = acceleration and v = fuel exit. m was given as 50,000 kg + 2000 Kg = 52,000 kg (without fuel) v was calculated from equation earlier with sqrt((2 * (P2  P1))/p), with given pressures and density. This came out as 2500 m/s
So... q = (52,000 * 37) / 2500 = 770 kg/s
From this the Thrust (T) = qv = (770 * 2500) = 1,925,000 N (It was assumed the pressures were the same so T = qv)
Calculating the total mass of the rocket on launchpad needs the weight of the fuel which I presume is the mass loss rate (770 Kg/s) multiplied by the duration of fuel (300 secs) = 231,000 Kg
Therefore the force due to gravity is... F = ma = (231,000 + 52,000) * 9.8 = 2,773,400 N.
This is more than the thrust calculated earlier which would give a negative net thrust which cannot be correct. Can anyone see where I've gone wrong please?
Regards

 
Aug 4th 2015, 09:32 AM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

I think there is a problem with this:
Originally Posted by deeppurple247 Equation given for this was was q = ma / v where q is the mass loss rate, m = final mass of rocket, a = acceleration and v = fuel exit 
The value for a can't possibly be the acceleration of the rocket. Consider a rocket that is merely hovering, so that a=0: this formula would indicate it requires no fuel to do so, which clearly is not right. I'm thinking the value for a ought to have something to do with the acceleration of the exhaust gasses, not the acceleration of the rocket. Or perhaps the term "ma" should actually be F_thrust (perhaps in a previous calculation it is assumed that F_thrust = ma?). Can you show us where you got the equation q=mv/a?

 
Aug 4th 2015, 10:54 AM

#3  Senior Member
Join Date: Apr 2008 Location: Bedford, England
Posts: 668

I note that you seem to be assuming a constant acceleration.
Since the mass is changing (as you burn fuel) but your thrust is constant, the acceleration must be changing.
This implies that your initial value for q is wrong.
This is the value of q to accelerate the empty rocket at 37m/s/s not the full rocket.
The initial acceleration will be zero, and will rise linearly as the fuel is burned.
So to get to the escape velocity the average acceleration will be 37.3, the final acceleration will be higher.
__________________
You have GOT to Laugh !

 
Aug 4th 2015, 03:32 PM

#4  Senior Member
Join Date: Apr 2008 Location: Bedford, England
Posts: 668

I've played around with this a bit and I can't get it to work with the exit velocity you quote.
My calculations suggest it would require an exit velocity of just over 3300 m/s.
It's quite possible I've made a mistake, but I think there is a problem with your original numbers.
Perhaps someone else out there will point out what we are missing, but I can't see it.
Please let us know if you have a eureka moment.
ps
What are P1 and P2 in your equation for v, I guess p is indicating rho (for density)?
__________________
You have GOT to Laugh !
Last edited by MBW; Aug 4th 2015 at 03:34 PM.

 
Aug 5th 2015, 12:37 AM

#5  Junior Member
Join Date: Aug 2015
Posts: 2

MBW you're correct. The exit velocity is actually 3535.65 m/s, and yes p was ro for density. However this still means the force of thrust required (T = qv) is less than the force due to gravity.
q now equals (52000 * 37) / 3535.65 = 544.2 Kg/s (From q = (m * a) / v)
FYI the equation in the back of the booklet definitely says these values are:
m = mass of the rocket
a = acceleration of rocket
q = fuel mass ejection rate
v = fuel exit velocity
And so T = 3535.65 * 554.2 = 1,959,440 N
One thing that may be wrong is the initial calculation of the acceleration for the escape velocity. The question said...
Calculate the acceleration required by the rocket to reach the escape velocity (11,200 m/s) within 5 minutes (300 seconds). There is no other information given at this point.
So I presume it's assuming a constant acceleration of 37.3 m/s/s ?
Regards

 
Aug 5th 2015, 05:50 AM

#6  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,289

The problem is here: the equation q= ma/v is incorrect. The actual equation to use is T = F_thrust = qv, which gives q=T/v. It seems someone then set T = ma, to get q=ma/v. But the proper equation is sum of forces = ma, so that:
Sum of forces = T  mg = ma > qvmg = ma >
q= m(ag)/v
Try this and see if it doesn't work out better for you.

 
Aug 5th 2015, 08:35 AM

#7  Senior Member
Join Date: Apr 2008 Location: Bedford, England
Posts: 668

Velocity of the Rocket : V = at
Acceleration of Rocket : a = M/T g
T is the Thrust
g is the acceleration due to gravity.
Total Mass of Rocket : M = m + f
m is empty mass of rocket
f is mass of fuel
Mass of Fuel : f = F  qt
F is initial mass of fuel
q is the fuel mass ejection rate.
Thus
Velocity of the Rocket : V = [(m+Fqt)/T g]t
Thrust : T = qv
v is the fuel exit velocity.
__________________
You have GOT to Laugh !
Last edited by MBW; Aug 5th 2015 at 04:06 PM.

  Search tags for this page 
Click on a term to search for related topics.
 Thread Tools   Display Modes  Linear Mode  