Gravity affects

[Peter Schutze]

I was trying to design a 2300 like universe where gravity has more effects than 2300 canon's "stutterwarp does / doesnt work" and have run into a problem ....

how do you work out a star's gravitational effect ?

According to wikipedia the sun has roughly the same effect out to saturn according to "escape velocity" as the earth does at surface level ie to escape earth gravity you need about 11 metres / second velocity while at saturn you need about the same to escape the sun's gravity and leave the solar system

Unfortunately when I try to calculate gravit,y I get the right answers at the various planets (and sun's) surfaces but the gravity effect on the sun calculates at something negligible even at venus's orbital distance:
formula is apparently g=GM/r^2 where G=gravitational force constant, M=body mass and r=distance (radius) from surface of body

The basic change to stutterwarp I was planning was that there are several effect bands:
at 0.1g stutterwarp is barely capable of stationkeeping (like canon)
at 0.01g stutterwarp efficiency is the % of lightspeed (ie efficiency 1 is 3,000 km/s)
at 0.001 its efficiency*10 as a lightspeed percent (ie efficiency 1 is 30,000 km/s)
at 0.0001g its efficiency times lightspeed (ie LY per year)

so its only when you get way past the oort cloud that you get up to the canon efficiency=lightspeed per day

 

[Anders]

The gravity formula you have (GM/r^2) is correct, it describes the gravitational acceleration at that distance (multiply by the mass of the starship to get the total force on it). Escape velocity is however different, it is sqrt(2GM/r). But escape velocities are irrelevant for stutterwarps anyway.

I like the more fine-grained bands, they make sense. Canon sharp efficiency reductions have always seemed a bit arbitrary. These bands still have the same effect, but there are no artificial jumps. Perhaps you could even fit a smooth curve through them.

I wonder what happens in very empty interstellar (or intergalactic) space? My answer would be that in principle the speed just goes higher and higher but in practice the drive has a maximum speed set by how quickly the drive can oscillate. Since this frequency is limited by the finite speed of electromagnetic impulses across it, there is an upper limit to stutterwarp speed. Some people might be working on special drives that work more efficiently over interstellar distances than in-system (or vice versa).

Note 1: It is interesting that the drive is sensitive to the gravitational acceleration, since that implies it is affected by spacetime curvature. Discharge again needs spacetime curvature, not low potential energy. I guess it should be obvious from the Einstein field equations, but I can't see it

Note 2: There seems to be a way of using the stutterwarp to get energy from nothing. Go from a place of low gravitational potential like close to the Earth to a place with high potential far away. Let gravity drag you back: this provides kinetic energy that can be used to do work. Now you can repeat the process again and again. In 2320 reality of course using the drive costs far more energy, but it is an ugly little loophole nevertheless. I would close it by saying that the potential energy differences actually have to be paid by the drive: jumping out of a deep gravity well actually requires slightly more energy than going in. This is a few kW, mostly noticeable for ships with small MHD plants.

 

[Peter Schutze]

>so its only when you get way past the oort cloud that you get up to the canon efficiency=lightspeed per day

this bugged me so much I finally broke down and asked a maths professor friend to check my work last night and the original formulae were correct ..... I stuffed something in the later work so 0.0001g of the sun is actually within jupiter's orbit not way out in the oort so Im going to have to halve each range band to get the effect I want

he also pointed out that things like the 100 diameter limit in normal traveller only works if the planet's density = earth or its not gravity related contrary to the examples. For the moon it would be 40 radii and for mars about 60 radii while the sun, the equivalent g effect would be felt at several hundred radii .... almost to earth orbit

conversely these (astronomically speaking) tiny distances for planets mean that when plotting a jump into a system only the star really matters .... just dont jump into a planetary orbit

 

[GJD]

I was trying to design a 2300 like universe where gravity has more effects than 2300 canon's "stutterwarp does / doesnt work" and have run into a problem ....

how do you work out a star's gravitational effect ?

According to wikipedia the sun has roughly the same effect out to saturn according to "escape velocity" as the earth does at surface level ie to escape earth gravity you need about 11 metres / second velocity while at saturn you need about the same to escape the sun's gravity and leave the solar system

Unfortunately when I try to calculate gravit,y I get the right answers at the various planets (and sun's) surfaces but the gravity effect on the sun calculates at something negligible even at venus's orbital distance:
formula is apparently g=GM/r^2 where G=gravitational force constant, M=body mass and r=distance (radius) from surface of body

The basic change to stutterwarp I was planning was that there are several effect bands:
at 0.1g stutterwarp is barely capable of stationkeeping (like canon)
at 0.01g stutterwarp efficiency is the % of lightspeed (ie efficiency 1 is 3,000 km/s)
at 0.001 its efficiency*10 as a lightspeed percent (ie efficiency 1 is 30,000 km/s)
at 0.0001g its efficiency times lightspeed (ie LY per year)

so its only when you get way past the oort cloud that you get up to the canon efficiency=lightspeed per day

I think, and I'm quite rusty at this, that your figures might be a bit off. The universal gravitation equation gives the force acting on a body, measured in Newtons. To get the acceleration, you need to divide the force by the mass of the object, giving you your acceleration due to gravity. Divide this acceleration by by 9.8ms^-1 to get the figure in g's.

Also I believe the radius is measured from the centre of mass of an object, not the surface.

G.