For the time and distance calculations:
This ignores the effects of gravity from Tanar's star. Since each of the planet, OLH and TPoT are at about the same distance from the star, they are all affected (accelerated) almost identical amounts, in almost identical directions. That does not change anything for matching velocity.
I also ignore the effect of Tanar's gravity on both OLH and TPoT. See below for my calculations on how small that affect would be. I leave the planet as an obstacle to avoid, delaying TPoT a little. When dealing with ships using Thruster plates (continuous boost), even close by planets have little effect. The Traveller books ignore that in provided tables for simplification. So do I.
Times below are shown in seconds, then converted to more readable values. Time 0 is when OLH exited jump, and promptly lost power. Velocities are shown in km/sec. Distances are shown in thousands of km (rounded). Positive distances are from the planet (Tanar) toward OLH's jump exit point. Negative distances are past Tanar. That is not a straight line, due to the course being bent by the close pass through the gravity well, but that does not matter for the calculations. It is only a distance without a direction. 1G acceleration approximated to 10 m/s^2.
Timeline
Time | OLH | TPot | |
---|
Sec | hh:mm:ss | km*1000 | km/s | km*1000 | km/s | Note |
---|
0 | 00:00:00 | 1146 | 80 | 0 | 0 | Jump Exit |
600 | 00:10:00 | 1098 | 80 | 0 | 0 | report from traffic control |
2100 | 00:35:00 | 1026 | 80 | 0 | 0 | TPoT gets underway at 2G |
4663 | 01:17:43 | 960 | 80 | 0 | 0 | TPoT gets clear of the planet |
7491 | 02:04:51 | 640 | 80 | 80 | -57 | TPoT reverses direction |
10320 | 02:52:00 | 320 | 80 | 160 | 0 | TPoT stopped relative to Tanar |
14320 | 03:58:40 | 0 | 80 | 0 | 80 | TPoT matches velocity with OLH |
16120 | 04:28:40 | -144 | 80 | -144 | 80 | TPoT dock and rig for towing; start 1G decel |
24120 | 06:42:00 | -464 | 0 | -464 | 0 | TPoT and OLH stopped relative to Tanar |
30932 | 08:35:32 | -232 | -68 | -232 | -68 | TPoT and OLH max v back toward Tanar |
37744 | 10:29:04 | 0 | 0 | 0 | 0 | TPoT and OLH stopped above starport |
39544 | 10:59:04 | 0 | 0 | 0 | 0 | TPoT and OLH land |
2G for 4000 seconds is 80km/s and 160000km travelled, while 80km/s moves 320000km
Tweaking the delay before they get underway, and how long it takes to get clear of the planet will adjust whether the rendezvous is a bit before or a bit after OLH's closest approach to the planet.
1G for 8000 seconds to slow from 80km/s moving 320000km in the process, when TPoT is stopping OLH.
For more detailed gravity calculations:
Mentioned previously,
F
grav = ( G * m
1 * m
2 ) / d
2
That gives the "force" pulling the masses toward each other. Both masses will be affected. from f = m * a, shuffling gives a = f / m. The force is the same, so each mass has a different acceleration.
The acceleration due to gravity on the surface of a planet or celestial body depends on its mass and radius. The formula to calculate the acceleration due to gravity (g) is:
g = (G * M)/R
2
or
a = (G * m)/R
2
When one mass is negligible compared to the other, the small mass will be "accelerated" toward the larger mass due to gravity with that formula
Where:
- g (and a) is the acceleration due to gravity.
- G is the universal gravitational constant (approximately 6.67430 × 10-11 m3/kg-1 s-2
- M is the mass of the planet or celestial body.
- R is the distance from the centre of the planet to the point where you want to calculate the acceleration due to gravity
Reorganizing,
(a * R
2) / G = m
For Tanar:
- a = 0.88g (.88 of a standard one earth gravity) = 8.8 m/s2
- R = 5600km (half of diameter of 11200km) = 5600000m
So
m
Tanar = (8.8 * 5600000
2 / G = about 4.13*10
24 Kg
We could now plug that back into
a = (G * m)/R
2
to calculate the gravitational acceleration at a different distance (R) from the centre of the planet. However, instead of working with big numbers again, the mass expression from the surface gravity can be plugged in instead of the calculated value.
a = (G * [(8.8 * 5600000
2 / G)]/R
2
"G" is in that twice, and cancels out
a = (8.8 * 5600000
2)/R
2 m/sec
2
a = 275968000000000 / R
2; were R is in meters. Converting
a = 275968000 / R
2; where R is in Km.
To test that, the surface is 5600Km (half the diameter) from the centre of the planet, so
a
surface = 275968000 / 5600
2 = 275968000 / 31360000 = 8.8 m/sec
2 = .88g
At that 50000Km exclusion limit
a = 275968000 / 50000
2 = 0.11 m/sec
2 or 0.011g
At the jump limit where OLH came out,
a = 275968000 / 1097600
2 = 0.00023 m/sec
2 or 0.000023g
I do not have the equation handy that would change
v
1 = v
0 + a*t
into an equation that handles "a" changing from 0.00023 to 0.11 m/sec
2 over the course of 3.5 hours (I think that needs an Integral), but even using the final (maximum) acceleration for the whole time:
v
1 = 80000 + 0.11*12520 = 81377.2
So the change in speed due to Tanars gravity is negligible. I ignore it, as do the transit time tables in the Traveller books.
This message was last edited by the player at 01:06, Tue 26 Sept 2023.