Joe wrote:

>Okay, here's an example of a distortion. These numbers seem familiar...
>Ah, yes - here it is on page 98 in my copy of Zubrin's "Entering Space",
>almost word for word. Trouble is, are these numbers accurate? For
>reference, let me point you to
>http://www.pma.caltech.edu/~chirata/deltav.html , which I found on my
>first Google search for '"Delta V" Mars Moon'.
>

Cool table, really.

But looking a little closer at this table I wonder: why does it say
delta-v Earth to LEO is 9.7 when LEO speed is 7.9 km/s?

The most considerate answer would be that it includes the delta-v
equivalent to go up to LEO-height, which they say is 400 km. And maybe
it includes losses due to air drag and fighting gravity during the
launch. While the last two depend on the space craft and its ascend
trajectory, the height part can be approximated via the necessary
energy: E/m = 400000m*10 m/s^2 which is equivalent to a speed of
v=sqrt(2*40000)m/s = 2.8 km/s. Now this would already give 7.9+2.8 =
10.7 km/s. Then we could argue gravity at 400 km is already a little
reduced: E = m*g*R^2*(1/R-1/(R+h)), gives E/m =
9.81m/s^2*6380^2km^2*(1/6380km - 1/(6380km+400km)) = 9.81m/s^2*376km =
3.7 MJ/kg or a speed of 2.7 km/s, so reduced gravity is not really
relevant here.

So whatever is computed here, it doesn't seem to be for a 400 km LEO!

Let's try the other way round: escape velocity is 11.2 km/s. Summing
up the delta-vs of the table from earth to escape (why is it called
c3=0?) gives 1.7 km/s more than escape velocity. Is this a reasonable
price to punch through atmosphere and fight gravity?
Say ascend takes 8 minutes, like the space shuttle or Saturn V. During
that time gravity has to be fighted less and less. The exact curve is
specific to the space craft, so for a simple estimate let's say its
equivalent to fighting full gravity for 4 minutes. This would be
v=9.81m/s^2*4*60s = 2.4 km/s, which is considerably larger than 1.7
km/s and its still missing air drag.

But look at the moon. There the sum of delta-v adds up to 2.3 km/s,
which is effectively moon escape velocity.

Seeing this I would be inclined to say this table is ignoring any
delta-v equivalent needed during launch from ground to low orbit and
has at least one typo.

But wait! How much would a spacecraft have to fight moon gravity?
Say 3g acceleration is used, 1.6 km/s takes 1600/30=53 seconds.
During this time moon accelerates the ship by much less than
1/6*9.81*53 = 90 m/s. This is so small a number it wouldn't show up in
this table.

Incredible how much easier it is to launch from the moon than from earth!

Mars sum to c3=0 is 5.5 km/s. What is mars escape? I get 5040 m/s from
the numbers in my astronomy book. 450 m/s is a reasonable penalty
while launching. On the other hand side, 400 m/s could also be missing
from the sum due to rounding errors.

So finally I still don't know where this 9.7 km/s number comes from.
Any suggestions? Do I miss something?

Best,
Axel

P.S.: as far as I see this problem does not affect any argumentation
previous posters did do based on this table.

> From: Axel Walthelm [mailto:Axel.Walthelm@...]
> So finally I still don't know where this 9.7 km/s number comes from.
> Any suggestions? Do I miss something?

I'm betting on typo & occum's razor.. ;)

ANTIcarrot.

ANTIcarrot wrote:

>>From: Axel Walthelm [mailto:Axel.Walthelm@...]
>>So finally I still don't know where this 9.7 km/s number comes from.
>>Any suggestions? Do I miss something?
>>
>>
>
>I'm betting on typo & occum's razor.. ;)
>
>ANTIcarrot.
>

Hi group,
for some reasons I didn't get back a copy of my original posting.
Does anybody experience similar problems? Or is the problem on my side?

Apart from that I got an interesting response from the author of the
delta-v table: