Gravity -- how do I work it out?

[Fitz]

Assume I have a ship which is capable of accelerating at 4G (40m/sec sq.) in a rapidly degrading orbit around a star with a surface gravity of 600m/sec sq.

 

Is there an easy way for a mathematical dullard such as myself to work out how close I'd be able to approach the star and still be able to get away? I'm not necessarily looking for absolute accuracy here, so much as a rule I can apply which would be good enoughh for space opera purposes.

[tkdguy]

You can calculate the force of attraction between the star and the ship by the equation F = GM/d^2. G is the universal constant of gravitation (6.67x10^-11Nm^2/kg^2), M is the star's mass (the ship's mass is negligible compared to the star unless it's the size of a giant planet), and d is the distance. Note that the inverse square law applies here, ie, if your ship is twice as far from the star than another ship, the star's force of attraction on your ship is 4 times weaker than on the other ship.

[Outsider]

Yup, yup. Inverse square law.

 

The two main pieces of data you need about the star are its radius and its "surface" gravity.

 

 

The sun, for example, has a radius of about 700,000 km, and a surface gravity of about 28 Gs.

 

Your ship could only hover just beyond 1.15 million km of the surface without being doomed. (((28/4)^1/2 * 700,000 km)-700,000)

 

It could, of course, get significantly closer if it came in with a significant velocity on a parabolic approach.

[Drakkenkin]

Someone please correct me if I'm wrong but here is how you would do it by math:

 

Givens: The ship starts with max acceleration and flies in a 'strait' path away from the Sun.

 

First the variables:

m = mass of smaller object (ship) in [kg]

M = mass of larger object (Sun) [kg]

G = Gravitational Constant [Nm^2/kg^2]

r = distance from center of the larger object to center of smaller object in [m]

Rs = radius of the Sun in [m]

a = current acceleration of ship

 

The Numbers for this equation:

m = need

M = 1.991*10^30 [kg]

G = 6.673*10^-11 [Nm^2/kg^2]

r = unknown

Rs = 6.96*10^8[m]

sum of all forces = ma

a = 40 [m/s^2]

 

Second the equation with its units:

(m[kg] * M[kg] * G[Nm^2/kg^2]) / (r[m])^2

 

Third, calculate the equation with just the units:

1[N] = 1[(kg * m) /s^2

([kg] * [kg] * [Nm^2/kg^2]) / [m^2]

= [Nm^2] / [m^2]

= [N]

 

Forth the calculations:

-1) To find the minimum distance a ship can get to the Sun and still get away with max acceleration we need to know the mass of the ship. It’s sad but true.

-2) set the equation to equal ( 40 [m/s^2] * m [kg]

 

Lets say in this example that the ship’s mass is 5000 [kg]. To find a mass that works for you, you might want to find out how much mass another object has and compare its size to the size of your ship. (EX: a combat ship may have the same mass per [m] as a tank while a racing ship may have a mass more in common with a car.) Anyway... on with the calculations,

 

Because you want to know how close the ship can get to the Sun we will leave 'r' as an unknown, and set the equation equal to it. Then we just solve the equation.

 

(5000[kg] * 1.991*10^30[kg] * 6.673*10^-11 [Nm^2/kg^2]) / (r[m])^2 = 5000[kg] * 40 [m/s^2]

 

=> ( 6.643*10^23[N m^2] ) / (r[m])^2 = 200000 [N]

=> ( 6.643*10^23[N m^2] ) / (200000 [N]) = (r[m])^2

=> 3.32*10^18 [m^2] = (r[m])^2

=> SQRT (3.32*10^18 [m^2]) = SQRT (r[m])^2

=> 1.82*10^9[m] = r[m]

 

That means that the ship could technically get with in about 1,820,000,000 [m] of the center of the sun OR (1.82*10^9[m] -Rs) 1,130,000,000 [m] from the surface of the sun. Just for a loose perspective, it's a good bit closer than the closest planet (Mercury) that's at 5.79*10^10 [m]!

 

SO, you say you want this to be an easy calculation that you can use over and over again. Well just plug in 'm', 'a', 'M' into the following equation, and grind out what r is:

 

SQRT ( M[kg] * 6.673*10^-11 [Nm^2/kg^2]) / (a[m/s^2]) = r[m]

 

Then subtract the radius of the Sun, and you will have the closest they could get (in meters) and still get away using max acceleration.

 

Drakkenkin

 

PS: I'm sorry, I just took a look at you original post and saw that you wanted it for a Sun with a surface g = 600 [m/s^2]. Hopefully this will help, but if I get the time I'll find a probable solution for your exact question.

[tkdguy]

Yes, your math is correct. I've only given it a cursory glance right now, but it looks about right. I think the ship's mass isn't too important when calcualting it because it is so tiny compared to the star, but you could argue it's very important to the people on board. I'm not fully awake just yet, so I'll take a closer look when I am.

[Drakkenkin]

Thanks for taking a look at all.

 

Originally posted by tkdguy I think the ship's mass isn't too important when calculating it because it is so tiny compared to the star

 

No, your right about not needing the mass of the ship. One, it is to small in comparison to the sun to make any noticeable difference in the solution and two it cancels out when you solve for r.

 

I'll make that correction in the last equation now. I was trying to stay to the original formula so people could use it for other heavenly bodies, but for solving for r I think it would make it more simple for people to use.

 

Drakkenkin

[tkdguy]

Nice to know. It would be very embarassing for someone with an astronomy degree and experience teaching physics to get that one wrong.

[Drakkenkin]

Sorry if I offended you in any way. I thought that you were not sure, because you posted,

Originally posted by tkdguy

I think the ship's mass isn't too important ... I'm not fully awake just yet, so I'll take a closer look when I am.

not that it was your opinion. My bad.

 

Do you currently teach at one of the local bay area schools?

 

Drakkenkin

[tkdguy]

No, I wasn't offended at all. Don't worry. I just haven't done it in a while, and I'm getting rusty. I teach mostly math right now at a special program. when I taught physics it was at another special program.

 

The reason I couldn't take a very good look at your post earlier today is because I had this strange feeling in my head. It wasn't exactly a headache, and it wasn't exactly dizziness. It could have been vertigo. I was at the computer, and I felt the need to lie down. Weird feeling.

[Drakkenkin]

Glad to hear I didn't offend you. I hope your feeling better now and I hope you don't have to experience it again!

 

Drakkenkin

[tkdguy]

Thanks. Me too.

[TaxiMan]

Technically, if a ship has 4G of acceleration, it can be arbitrarily close to anything short of a black hole and escape. That's assuming no drag, which isn't exactly true in the corona of a star or some such.

 

The acceleration has to be sideways to the gravity pull, so the ship will be speeding up continuously. Forever. Eventually it will reach orbital speed & stop falling. From then on it will climb higher until it can redirect some of that acceleration away from the star & escape.

 

If you only want to escape by going directly away from the star / whatever, then you can escape as long as the gravity at that point is less than 4G.

[Guest zarglif69]

Originally posted by Drakkenkin

Someone please correct me if I'm wrong but here is how you would do it by math:

 

Givens: The ship starts with max acceleration and flies in a 'strait' path away from the Sun.

 

First the variables:

m = mass of smaller object (ship) in [kg]

M = mass of larger object (Sun) [kg]

G = Gravitational Constant [Nm^2/kg^2]

r = distance from center of the larger object to center of smaller object in [m]

Rs = radius of the Sun in [m]

a = current acceleration of ship

 

The Numbers for this equation:

m = need

M = 1.991*10^30 [kg]

G = 6.673*10^-11 [Nm^2/kg^2]

r = unknown

Rs = 6.96*10^8[m]

sum of all forces = ma

a = 40 [m/s^2]

 

Second the equation with its units:

(m[kg] * M[kg] * G[Nm^2/kg^2]) / (r[m])^2

 

Third, calculate the equation with just the units:

1[N] = 1[(kg * m) /s^2

([kg] * [kg] * [Nm^2/kg^2]) / [m^2]

= [Nm^2] / [m^2]

= [N]

 

Forth the calculations:

-1) To find the minimum distance a ship can get to the Sun and still get away with max acceleration we need to know the mass of the ship. It’s sad but true.

-2) set the equation to equal ( 40 [m/s^2] * m [kg]

 

Lets say in this example that the ship’s mass is 5000 [kg]. To find a mass that works for you, you might want to find out how much mass another object has and compare its size to the size of your ship. (EX: a combat ship may have the same mass per [m] as a tank while a racing ship may have a mass more in common with a car.) Anyway... on with the calculations,

 

Because you want to know how close the ship can get to the Sun we will leave 'r' as an unknown, and set the equation equal to it. Then we just solve the equation.

 

(5000[kg] * 1.991*10^30[kg] * 6.673*10^-11 [Nm^2/kg^2]) / (r[m])^2 = 5000[kg] * 40 [m/s^2]

 

=> ( 6.643*10^23[N m^2] ) / (r[m])^2 = 200000 [N]

=> ( 6.643*10^23[N m^2] ) / (200000 [N]) = (r[m])^2

=> 3.32*10^18 [m^2] = (r[m])^2

=> SQRT (3.32*10^18 [m^2]) = SQRT (r[m])^2

=> 1.82*10^9[m] = r[m]

 

That means that the ship could technically get with in about 1,820,000,000 [m] of the center of the sun OR (1.82*10^9[m] -Rs) 1,130,000,000 [m] from the surface of the sun. Just for a loose perspective, it's a good bit closer than the closest planet (Mercury) that's at 5.79*10^10 [m]!

 

SO, you say you want this to be an easy calculation that you can use over and over again. Well just plug in 'm', 'a', 'M' into the following equation, and grind out what r is:

 

SQRT ( M[kg] * 6.673*10^-11 [Nm^2/kg^2]) / (a[m/s^2]) = r[m]

 

Then subtract the radius of the Sun, and you will have the closest they could get (in meters) and still get away using max acceleration.

 

Drakkenkin

 

PS: I'm sorry, I just took a look at you original post and saw that you wanted it for a Sun with a surface g = 600 [m/s^2]. Hopefully this will help, but if I get the time I'll find a probable solution for your exact question.

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