Manned Mission to Mercury

Jim Davis wrote:
"Kaido Kert" wrote ...


My crude calculation shows, that 1 kg at 400kmh orbit, holds about
370Mjoules of energy, relative to one standing on earth. Thats about
100kWH, which doesnt cost a whole lot of money, when you get it from
power grid.


You misplaced a decimal point somewhere. It's about 33 MJ or 9.2 kWh.

In other words the energy it takes to leave a 100 watt bulb on for 3.8 days?


Jim Davis




How do you calculate this? Maybe it's in my Mueller Bates and White book.

--
Hop David
http://clowder.net/hop/index.html

On Thu, 30 Oct 2003 11:17:12 -0700, in a place far, far away, Hop
David made the phosphor
on my monitor glow in such a way as to indicate that:


Did calculations leaving LMO for Hohmann:

Escape velocity at 100 km: 4.16 km/sec

What is the relevance of escape velocity?

Vmercury - Vhohmannperihelion: 9.61 km/sec

hyperbolic excess velocity
sqrt(4.16^2 + 9.61^2) km/sec 10.48 km/sec

I've no idea what you're doing here.

Parking orbit velocity 2.95 km/sec

Insertion burn (10.48-4.16) km/sec 7.53 km/sec

I can't figure out why you're doing a ninety-degree vector addition of
those two velocities.

Take the 9.61 km/s to be your velocity at infinity. Using
conservation of energy, compute the velocity you need at apoapsis to
get that velocity at infinity. That's the delta V you need to land on
(or take off from) Mercury. If you want the delta V to just
circularize, subtract from that velocity the local circular.

--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org

"Extraordinary launch vehicles require extraordinary markets..."
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"Hop David" wrote...

You misplaced a decimal point somewhere. It's about 33 MJ or 9.2 kWh.

In other words the energy it takes to leave a 100 watt bulb on for 3.8
days?

Yes.

How do you calculate this? Maybe it's in my Mueller Bates and White book.

Total Energy = Kinetic energy + Potential Energy = g0 * R^2 / 2 / r + g0 *
R * ( 1 - R / r)

where

g0 = 9.81 m/s^2
R = radius of earth
r = distance of satellite from center of earth

Jim Davis

Hop David wrote:

Jim Davis wrote:
"Kaido Kert" wrote ...


My crude calculation shows, that 1 kg at 400kmh orbit, holds about
370Mjoules of energy, relative to one standing on earth. Thats about
100kWH, which doesnt cost a whole lot of money, when you get it from
power grid.


You misplaced a decimal point somewhere. It's about 33 MJ or 9.2 kWh.

In other words the energy it takes to leave a 100 watt bulb on for 3.8 days?


Jim Davis



How do you calculate this? Maybe it's in my Mueller Bates and White book.

The total energy is the sum of the potential energy and kinetic energy:

E = mgh + 1/2(mv^2)
= (1)(9.81)(400)(1,000) + (.5)(1)(7,669)^2
= (3.924)(10^6) + (29.41)(10^6)
= 33.33 MJ


--
Hop David
http://clowder.net/hop/index.html

On Thu, 30 Oct 2003 21:27:40 GMT, in a place far, far away, Dick
Morris made the phosphor on my
monitor glow in such a way as to indicate that:

The total energy is the sum of the potential energy and kinetic energy:

E = mgh + 1/2(mv^2)
= (1)(9.81)(400)(1,000) + (.5)(1)(7,669)^2
= (3.924)(10^6) + (29.41)(10^6)
= 33.33 MJ

That's an approximation. You shouldn't use the flat-earth version,
though the error is relatively small for LEO.

--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org

"Extraordinary launch vehicles require extraordinary markets..."
Swap the first . and @ and throw out the ".trash" to email me.
Here's my email address for autospammers:

Rand Simberg wrote:

On Thu, 30 Oct 2003 21:27:40 GMT, in a place far, far away, Dick
Morris made the phosphor on my
monitor glow in such a way as to indicate that:

The total energy is the sum of the potential energy and kinetic energy:

E = mgh + 1/2(mv^2)
= (1)(9.81)(400)(1,000) + (.5)(1)(7,669)^2
= (3.924)(10^6) + (29.41)(10^6)
= 33.33 MJ

That's an approximation. You shouldn't use the flat-earth version,
though the error is relatively small for LEO.

True, it neglects the variation of gravity with altitude. The potential
energy is about 5% off.
--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org

"Extraordinary launch vehicles require extraordinary markets..."
Swap the first . and @ and throw out the ".trash" to email me.
Here's my email address for autospammers:

In article , Hop David wrote:

You misplaced a decimal point somewhere. It's about 33 MJ or 9.2 kWh.

In other words the energy it takes to leave a 100 watt bulb on for 3.8 days?

Looks about right.

For an estimation of the "counter price" of that,
http://fox.rollins.edu/~jsiry/93price.htm may be of interest. Of course,
it's a pretty nebulous calculation anyway... g

(By the time you hit anything even remotely resembling
pocket-money-per-kilo prices to orbit, I wouldn't want to be predicting
what electricity prices will look like...)

--
-Andrew Gray

Rand Simberg wrote:
On Thu, 30 Oct 2003 11:17:12 -0700, in a place far, far away, Hop
David made the phosphor
on my monitor glow in such a way as to indicate that:



Did calculations leaving LMO for Hohmann:

Escape velocity at 100 km: 4.16 km/sec


What is the relevance of escape velocity?

Vinfinity^2 = Vburnout^2 - Vesc^2

paraprhasing equation 1.10-5 in _Fundamentals of Astrodynamics_ by Bate,
Mueller and White.



Vmercury - Vhohmannperihelion: 9.61 km/sec

hyperbolic excess velocity
sqrt(4.16^2 + 9.61^2) km/sec 10.48 km/sec


I've no idea what you're doing here.


Parking orbit velocity 2.95 km/sec

Insertion burn (10.48-4.16) km/sec 7.53 km/sec


I can't figure out why you're doing a ninety-degree vector addition of
those two velocities.

That's what I thought when I first saw above equation 1.10-5: Vburnout
the hypotenuse, Vescape & Vinfinity the legs of a right triangle.

However I haven't been able to find this right triangle in the sketches
I've made.


Take the 9.61 km/s to be your velocity at infinity. Using
conservation of energy, compute the velocity you need at apoapsis

You mean periapsis?

to
get that velocity at infinity. That's the delta V you need to land on
(or take off from) Mercury.


Yes, that seems to be the approach Bate Mueller and White used. Just
preceding equation 1.10-4 was something like this:

E = V^2/2 - Gm/r

E = Vburnout^2/2 - Gm/rburnout

E = Vinfinity^2/2 - 0

Vinfinity^2/2 = Vburnout^2/2 - Gm/rburnout

Vinfinity^2 = Vburnout^2 - 2 Gm/rburnout

Vinfinity^2 = Vburnout^2/2 - Vesc^2

If you want the delta V to just
circularize, subtract from that velocity the local circular.


Yes, I subtracted the circular orbit velocity from Vburnout.

--
Hop David
http://clowder.net/hop/index.html

On Fri, 31 Oct 2003 08:32:43 -0700, in a place far, far away, Hop
David made the phosphor
on my monitor glow in such a way as to indicate that:


Take the 9.61 km/s to be your velocity at infinity. Using
conservation of energy, compute the velocity you need at apoapsis

You mean periapsis?

Yes, sorry.

to
get that velocity at infinity. That's the delta V you need to land on
(or take off from) Mercury.


Yes, that seems to be the approach Bate Mueller and White used. Just
preceding equation 1.10-4 was something like this:

E = V^2/2 - Gm/r

E = Vburnout^2/2 - Gm/rburnout

E = Vinfinity^2/2 - 0

Vinfinity^2/2 = Vburnout^2/2 - Gm/rburnout

Vinfinity^2 = Vburnout^2 - 2 Gm/rburnout

Vinfinity^2 = Vburnout^2/2 - Vesc^2

If you want the delta V to just
circularize, subtract from that velocity the local circular.


Yes, I subtracted the circular orbit velocity from Vburnout.

It may be mathematically equivalent, but I sure couldn't follow it.
BM&W is not my favorite book. It's much stronger on orbit
determination than on trajectory planning.

--
simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole)
interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org

"Extraordinary launch vehicles require extraordinary markets..."
Swap the first . and @ and throw out the ".trash" to email me.
Here's my email address for autospammers:

Rand Simberg wrote:
On Fri, 31 Oct 2003 08:32:43 -0700, in a place far, far away, Hop
David made the phosphor
on my monitor glow in such a way as to indicate that:



Take the 9.61 km/s to be your velocity at infinity. Using
conservation of energy, compute the velocity you need at apoapsis

You mean periapsis?


Yes, sorry.


to
get that velocity at infinity. That's the delta V you need to land on
(or take off from) Mercury.


Yes, that seems to be the approach Bate Mueller and White used. Just
preceding equation 1.10-4 was something like this:

E = V^2/2 - Gm/r

E = Vburnout^2/2 - Gm/rburnout

E = Vinfinity^2/2 - 0

Vinfinity^2/2 = Vburnout^2/2 - Gm/rburnout

Vinfinity^2 = Vburnout^2 - 2 Gm/rburnout

Vinfinity^2 = Vburnout^2/2 - Vesc^2


If you want the delta V to just
circularize, subtract from that velocity the local circular.


Yes, I subtracted the circular orbit velocity from Vburnout.


It may be mathematically equivalent, but I sure couldn't follow it.

?

E = V^2/2 - Gm/r

is elementary, no?

Then what comes after seems perfectly straightforward.

BM&W is not my favorite book. It's much stronger on orbit
determination than on trajectory planning.


What's your favorite book(s) on trajectory planning?

Am still hoping someone will check my figures. With self study I lack
instructors grading homework/tests. So it's harder for me to tell if I'm
harboring misconceptions.

--
Hop David
http://clowder.net/hop/index.html