[geocentrism] Re: New paper
- From: "Robert Bennett" <robert.bennett@xxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Wed, 13 Jun 2007 09:31:58 -0400
Neville,
From the Artemis site; F= 0.015 N (the constant thrust of Rita) and
payload plus propellant mass ~ 2000 kg. So Acc = F/m = 7.5 *10-6 m/s2
For const acc, the final velocity is vf = at.
Suppose Rita is aimed radially down for 12 hours, then up for 12 hours, for
340 days. After each day it will stop rising.
After 12 hours :
vf = at = 7.5 *10-6 m/s2 * 12hrs * 3600sec/hr = .32 m/s
The average velocity v is half of this, v = .16 m/s. (for both the first and
last 12 hours).
The distance traveled in 12 hrs is d=vt = .16 m/s * 12hr * 3600 sec/hr ~= 7
km; each day Artemis rises ~ 14 km. (The site says 15 km/day)
14 km/day* 340 days ~= 4800 km = 4.8 Mm. The site gives the change in
altitude as (3.6 - 3.1) Mm = 5 Mm.
The simple alternation of thrust direction every half day explains the
motion described for Artemis, using the 2nd law and the kinematic equations
for constant acceleration.
Robert
-----Original Message-----
From: geocentrism-bounce@xxxxxxxxxxxxx
[mailto:geocentrism-bounce@xxxxxxxxxxxxx]On Behalf Of Neville Jones
Sent: Tuesday, June 12, 2007 4:57 PM
To: geocentrism@xxxxxxxxxxxxx
Subject: [geocentrism] Re: New paper
Robert,
I amended the computer program and ran it through again, using 171.5 days
for thrust directed towards the World and 171.5 days for thrust directed
away from the World.
Assuming Eqs. 9 and 11 in "Geosynchronous satellites in a geostationary
universe" hold, then we obtain
v = 0 m/s
s = 1.062 x 10^9 m
This would result in a satellite over 30.5x higher than the altitude of the
'parking orbit'.
I will now try to obtain further details.
Neville.
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