snipped-for-privacy@juno.com wrote:
>
> dave.harper wrote:
>
>> snipped-for-privacy@juno.com wrote:
>>
>>
>>>So, ~95 percent off the energy/work goes into just getting off the
>>>ground and up to an altitude where an orbital path can be reached.
>>
>>I'm going to disagree with that statement. The potential energy
>>required to get to an altitude of 100 miles (neglecting the decrease in
>>gravitational acceleration) is:
>>
>>m

***g***h = 1,582,385 J/kg
>>
>>The kinetic energy of an object in orbit is:
>>
>>.5

***m***v^2 = 32,589,849 J/kg
>
>
> Yes, that is why I neglected the potential energy to make the
> calculation
> a bit more straight forward. I also took a mass to be VERY small, so
> that potential wouldn't be much of a factor.
>
>
>>The kinetic energy of an object in orbit is MUCH greater than the
>>potential energy needed to get there. If you've ever looked at a
>>launch schedule, you'll see that the vast majority of a rocket's burn
>>is excecuted well over the atmosphere. That's because it takes A LOT
>>more energy to get up to orbital velocity than it does to get up past
>>the atmosphere. Secondly, atmospheric drag only accounts for less than
>>about 30% of the total energy of a launch.
>
>
> Ummm the vast majority of a rockets burn is ued up reaching orbital
> altitude ... I do believe and I will check and post this, but the first
> stage of a Saturn V contains more fuel than the rest of the vehicle ..
> let me google the Sat V specs ... yes here it is ... the first stage
> burns the most propellant ... for the first 37 miles of the flight:
>
> Size: 111 m (363 ft)
> Payload to orbit: 129,300 kg (285,000 lb)
> Payload to Moon: 48,500 kg (107,000 lb)
> Manufacturer: Boeing Co. (prime)
> 1st stage: five F-1 engines
> Propellants: RP-1 (kerosene) and liquid oxygen
> Total thrust: 33,360,000 newtons (7,500,000 lb)
> Manufacturer: Rocketdyne
> 2nd stage: five J-2 engines
> Propellants: liquid hydrogen and liquid oxygen
> Total thrust: 5,560,000 newtons (1,250,000 lb)
> Manufacturer: Rocketdyne
> 3rd stage: one J-2 engine
> Thrust: 1,112,000 newtons (250,000 lb)
>

Nothing here backs up your claim that most of the fuel is in the first stage. I
am not saying it isn't true, but you make a claim then provide information which
you say backs it up. Except that it doesn't. A very poor way to win a debate.
Then there is the significant detail that the first stage does more than just
lift the rocket up. It also adds a significant fraction of the required orbital
velocity.

Not to mention that you started with an energy argument and have now switched
to mass. Yet the first stage of the Saturn V uses a low ISP fuel which of course
increases the mass in that stage.

Getting back to the claim that 95% of the energy expended is used

__ _just_ __to
reach orbital altitude rather than orbital velocity, perhaps an example will
illustrate the problems with that arguement.

Assume a SSTO ship to keep things simple. The Delta-V budget for LEO typically
given is 7.8km/s orbital velocity plus 1.5 to 2 km/s for drag and gravity
losses. Also assume that the rocket first expends fuel on the gravity/drag
losses and then does the orbital burn. This isn't true but it keeps things
simple and actually favors your arguement. Assume SSME exhaust velocity of 4500
m/s.

Mass left after 2 km/s burn:

m1 = m0 e^(-2,000/4500) = 0.64 m0

and after the 7.8km/s burn:

m2 = m1 e^ (-7,800/4,500) = 0.18 m1

36% of the initial launch mass is burned overcoming gravitational/drag losses.
82% of what is left or 52% of the initial launch mass is required for the
orbital velocity.

52% vs. 36%

I fail to see anything even vaguely resembling a 20 to 1 ratio.

If you used Saturn V first stage engines that have an ISP of 304s for the 2km/s
second burn, the mass fraction would be:

m1 = m0 e^( -2,000/ 304s * 9.8m/s/s) = 0.51

>
> So I am a bit confused by your statement above where you point out that
> the
> majority of the rocket's burn is above the atmosphere ... please
> elaborate.
>
>
>>>One way to think about this is through orbital and earth escape
>>>velocities: It requires a mass of any size/density to reach a speed of
>>>17,800 mph to stay in orbit at ~90 mi. Now for that very same mass, it
>>>requires 25,000 mph to have enough velocity to escape the Earth's
>>>influence entirely. One can clearly see that the speed needed in NOT
>>>doubled as its only 7,200 mph more. Why, because most of the work
>>>has ALREADY been done just getting to orbit.
>>
>>Actually, you're almost DOUBLING the kinetic energy an object by
>>accelerating it from 17,800mph to 25,000mph. Secondly, using orbital
>>velocity as a baseline isn't exactly realistic, since orbit is a VERY
>>high energy state.
>
>
> Well the Spaceship One had an orbit ... it just failed to put enough
> energy into its orbit to miss the Earth ...
>
>
>>When you're in orbit, you've overcome very little of the earth's
>>gravitational well, but you have a lot of kinetic energy. Think of it
>>this way: if you were to hold an object 500,000 miles over earth, then
>>drop it, it would hit the ground at over 7 miles per second. And
>>that's just from the conversion of gravitational potential energy to
>>kinetic energy.
>>
>
>
>
> Newton's law for Gravity - F = GMm/r*r as you leave the Earth the
> force of Gravity falls off VERY quickly. Remember, gravity is a WEAK
> force. If you don't believe this ... just stand up! Notice that the
> nuclear forces of your floor easily over come the force of gravity
> (your weight) pressing down upon it). Gravity is a very weak force in
> the universe.
>

The difference in the gravitational force from sea level to LEO is small. Mean
radius of the Earth is 3959 miles, assume an LEO altitude of 100 miles. So the
difference (assuming point sources) is 3959^2/ (3959+100)^2 or 95% of the sea
level value.

If gravity falls off so rapidly, why does the ESA claim that the force of
gravity at the ISS is 88% of that at sea level?

http://www.spaceflight.esa.int/users/file.cfm?filename úc-iss-pe-g

>
>>Hope that helps,
>>Dave
>
>
> I hope that helps too :)
>
> Lunarlos
>

--

David W. Schultz

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