[Kryptid]

What is the velocity of free fall of the ISS?

There is no one velocity of free fall in a vacuum. There is an acceleration, but not a velocity.

This proves the Newton gravity equation is false

No it doesn't. It proves that you have no idea how physics works.

since the mass should not be weightless.

Mass and weight are not the same thing. Try weighing a free-falling object on a free-falling scale in a vacuum. It won't measure anything because both will accelerate at the same rate.

[Janus]

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.

This entirely depends on how you measure "efficiency".
Let's take the Saturn V third stage for example.  It has a "fuel" load of 109500 kg. This consists of both liquid Hydrogen and liquid Oxygen.  Of this, it is the Hydrogen that provides the energy.   They are burned together to form water( H2O)  You use 2 kg of Hydrogen for 16 kg of Oxygen, so the Hydrogen makes up ~12167 kg of the load.   At 141.86e6 joules per kg, this gives an available energy of 1.73e12 joules. 
The rocket engine has an ISP 421 seconds, which equates to an exhaust velocity of 4130 m/sec.   At 4130 m/sec relative to the rocket engine, the total KE of the exhaust gases will be 9.339e 11 joules.   9.33e11/1.73e12 =.54. 
So if you rate your engine by what percentage of the energy released by burning the fuel goes into producing velocity for your exhaust and thus generating thrust, you get an efficiency of 54%.

[Janus]

. To put a tonne in orbit at 400km requires you to put roughly HALF the fuel at 200km.

No.  The Saturn V first stage only got the rocket up to 67 km of the 170 km LEO altitude.  Doing so, it used up ~80% of the total fuel load for the rocket.    So it only had to lift 20% of its fuel to under 40% of its targeted orbit  height.

Quit trying to guess at the answers.

[Janus]

Then why are astronaut in the ISS weightless?

The same reason that a ball dropped off of the Empire State Building is weightless: because they are in free fall.

What is the velocity of free fall of the ISS? This proves the Newton gravity equation is false since the mass should not be weightless. Right If it is falling (free fall) it must be falling downward which it is not. Pretty good there old chap. I was walking through the halls of Harvard and I thought of this.

It is "falling", in that its trajectory is always being curved to towards the Earth.  Acceleration is a change in velocity. Velocity is a vector which includes both magnitude and direction.  A body in a circular orbit is constantly changing its direction, but not its speed. This is still an acceleration At the ISS altitude, the acceleration due to gravity is ~8.66m/sec^2.  The orbital velocity of the ISS is  ~7.67 km/sec.  Thus it "falls" ~ 4.33 m as it travels forward 7.67 km.  This is a gentle curve*.  So gentle that the Earth curves out from under it just as fast. Look up Newton's cannon.

* To a close approximation. The fact that gravity acts towards the center of the Earth, and not just perpendicular to the initial velocity vector of the ISS, alters this value by some.

 

[Petrochemicals]

. To put a tonne in orbit at 400km requires you to put roughly HALF the fuel at 200km.

No.  The Saturn V first stage only got the rocket up to 67 km of the 170 km LEO altitude.  Doing so, it used up ~80% of the total fuel load for the rocket.    So it only had to lift 20% of its fuel to under 40% of its targeted orbit  height.

Quit trying to guess at the answers.

Or another way of thinking of it.

It has to lift the craft. The fuel required to go half way requires no extra fuel, but the other half of the fuel has to be lifted half way, thus requiring fuel itself to provide energy to be lifted, this itself requires fuel. anonanonandariston. Parabolic usage.

 The bigger the fuel load  the larger the coontainment, as was realised by Isombard kingdom brunel, and the coal powered craft he made. The problem is that the Ss Great eastern didnt need to float up a pepetual hill, where as rockets are.  The faster you can get them out of the atmosphere  the more efficient they become.

[Janus]

Why does not a penny in the ISS stick to the walls?

The mass of the ISS is 420,000 kg.  Even if all this mass were compacted into a point, an object held just 5 m away would take 59 min to reach it.  That's with all the gravity of the ISS puling in one direction.  Something inside the ISS would be surrounded by its mass, the gravity of which would be pulling it in all directions and for the most part canceling itself out. If the ISS were perfectly spherical with wall of uniform density, The pull of gravity would cancel out completely everywhere in the interior.  The ISS is a bit more asymmetric, So you are going to likely end up with some net pull in different points of the interior, but it will be extremely small.  Left alone for long enough, objects floating inside would eventually find resting spots, but not until a long , long time,  But even after that, just turning on the ventilation system would be enough to dislodge them from their resting spots.  The gravitational forces involved are just that small.

If you are asking why the penny doesn't stick to the "Earthside" wall, both the penny and ISS are "falling" along the same path due the same gravity.  There is no reason for the penny to fall any differently than the ISS. 
If you were in an elevator, and ti suddenly started to fall, between the time it started falling and hits the bottom of the shaft, you would float around inside the elevator just like the Astronauts do in the ISS. Both are examples of objects in free fall. You actually experience this partially in an elevator when it first starts downward (more so in fast elevators).  As the elevator accelerates to reach full descent speed, there is a brief period were you will feel lighter.  The downward acceleration of the elevator decreases weight you feel. If you normally weighed 160 lbs, and the elevator accelerated downward at 1 m/sec^2, then while the elevator was accelerating, you would only "weigh"  143.7 lbs ( this is what a spring or force reading scale would measure if you were standing on it.)

[Petrochemicals]

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuel-mass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.

From https://arxiv.org/pdf/1308.4869

[Janus]

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuel-mass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.

From https://arxiv.org/pdf/1308.4869

Since you did not include the specifics of what "arduous journey" was being contemplated here, its a bit hard to reply directly, but I am going to assume we are talking about an interstellar journey at some significant fraction of c

Fuel to payload ratios are determined by the rocket equation.

dV= Ve ln(MR)
where dV is the velocity change you get for your rocket
Ve is the exhaust velocity
ln(MR) is the natural log of the mass ratio ( fueled mass/dry mass) for the rocket.

If you need dV to be a significant fraction of the speed of light, there is a relativistic version.

To get the Mass ratio for any given dV, you rearrange to

MR= e^(dV/Ve)
where e is Eulers number = 2.718...

That being said, the quoted passage is neither here nor there as far as this thread is concerned, as we are not talking about sending craft to distant stars, but to the Moon.   Applying the above equation to the Saturn V rocket shows that it was within its capabilities to do the job for which it was intended.

[alright1234 (OP)]

What is the velocity of free fall of the ISS?

There is no one velocity of free fall in a vacuum. There is an acceleration, but not a velocity.

This proves the Newton gravity equation is false

How can there be an acceleration without a velocity?

[alright1234 (OP)]

Also, I have began my calculation of the efficiency of a rocket using the finial velocity of 938 m/s and weight of 50,000 kg and the rocket fuel energy of 2.4 x 10^7 J/kg. Watson jr. is computing and will produce the final number.

[Bored chemist]

Also, I have began my calculation of the efficiency of a rocket using the finial velocity of 938 m/s and weight of 50,000 kg and the rocket fuel energy of 2.4 x 10^7 J/kg. Watson jr. is computing and will produce the final number.

Why not do it correctly instead?

[alright1234 (OP)]

Show me the correct way.


The total fuel load is
109,500 +  456,100  + 2,160,000 = 2,725,600 kg

[alright1234 (OP)]

The kinetic energy of the CMSL propagating to the moon on a straight path is determined using the velocity of 938 m/s,



1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J........................................................................................1



The energy of 2,725,600 kg of fuel using the energy of a kilogram of rocket fuel (4.2 x 107 J/kg),



Fuel Energy = (mass fuel) x (energy of fuel) =  (2.72 x 106 kg)/(4.2 x 107 J/kg) =  1.142 x 1014 J....................2



Divided by kinetic of the CMSL and the energy of the fuel is,



Efficiency = (KE)/(enegy of fuel) =  (2.39 x 1010 J)/(1.142 x 1014) =  .0002 or .02 %........................................2



Multiply the efficiency by 100 times to normalize the effect of the earth's gravity forms an efficiency of free space of  2 %

[Kryptid]

Your calculations assume that all of the fuel energy went into accelerating the CMSL alone. It didn't. A significant portion went into accelerating the other stages of the vehicle as well. You need to take that into consideration.

[Petrochemicals]

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuel-mass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.

From https://arxiv.org/pdf/1308.4869

Since you did not include the specifics of what "arduous journey" was being contemplated here, its a bit hard to reply directly, but I am going to assume we are talking about an interstellar journey at some significant fraction of c

Fuel to payload ratios are determined by the rocket equation.

dV= Ve ln(MR)
where dV is the velocity change you get for your rocket
Ve is the exhaust velocity
ln(MR) is the natural log of the mass ratio ( fueled mass/dry mass) for the rocket.

If you need dV to be a significant fraction of the speed of light, there is a relativistic version.

To get the Mass ratio for any given dV, you rearrange to

MR= e^(dV/Ve)
where e is Eulers number = 2.718...

That being said, the quoted passage is neither here nor there as far as this thread is concerned, as we are not talking about sending craft to distant stars, but to the Moon.   Applying the above equation to the Saturn V rocket shows that it was within its capabilities to do the job for which it was intended.

I just thought you may want to go and tell them wahts what ?

[Janus]

The kinetic energy of the CMSL propagating to the moon on a straight path is determined using the velocity of 938 m/s,



1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J........................................................................................1



The energy of 2,725,600 kg of fuel using the energy of a kilogram of rocket fuel (4.2 x 107 J/kg),



Fuel Energy = (mass fuel) x (energy of fuel) =  (2.72 x 106 kg)/(4.2 x 107 J/kg) =  1.142 x 1014 J....................2



Divided by kinetic of the CMSL and the energy of the fuel is,



Efficiency = (KE)/(enegy of fuel) =  (2.39 x 1010 J)/(1.142 x 1014) =  .0002 or .02 %........................................2



Multiply the efficiency by 100 times to normalize the effect of the earth's gravity forms an efficiency of free space of  2 %

Where in the World did you pull the 938m/sec from?   It is nowhere near the Delta v required to get the CMSL to the Moon and back.    The CMSL left LEO moving at over 10,900 m/sec, and the energy required to do this also had to include the energy needed to climb to that altitude.   If you are going to include the fuel in the CMSL, then you need to add any velocity changes performed by engine usage to that figure.
You can't just go by the difference between initial and final velocities. As also stated above, you also need to take into account the effects of staging.

So for example, upon reaching the proper point near the Moon, the CMSL fired its engines in order to brake into Lunar orbit.  Upon leaving, it had to fire its engines to accelerate away from the Moon.  Even if there were no net difference between the velocity just before braking and just after acceleration, the total delta v you would calculate for the craft would be equal to the sum of the two speed changes made. (For example, if it had to brake by 1000 m/sec, and then later accelerate by 1000 m/sec, the total dv budget for the engine would be 2000 m/sec.

Another problem with your calculation is that that you apply the joules/kg figure to the entire "wet load" of the rocket.  So for example, when you say that the CMSL has "fuel" load of 109500 kg, this is not technically correct. It carries a combined mass of liquid Hydrogen and liquid Oxygen of 109500 kg. It is only the Hydrogen that the joules/kg figure applies to   The Hydrogen only makes up ~ 11.11% of that 109500 kg or ~12155 kg.  This is the mass you have to multiply by 4.2e7 joules/kg to get the energy content of the fuel, not the entire 109500 kg.

[yor_on]

Why does not a penny in the ISS stick to the walls?

Thinking of spin?

Otherwise I would go with Kryptid

[alright1234 (OP)]

The fact that NASA is using a halve circle trajectory to propagate to the moon using the gravitational force of the earth is senseless logic that cannot be justified since an astronaut is weightless in the ISS.

[Colin2B]

The fact that NASA is using a halve circle trajectory to propagate to the moon using the gravitational force of the earth is senseless logic that cannot be justified since an astronaut is weightless in the ISS.

Your logic is flawed, please take the time to learn some orbital physics. You could start with Hoheman Transfer Orbit.

[alright1234 (OP)]

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.