Orbital Velocity Question. Please Help!

[Der Anfang]

I'm trying to understand orbital velocity here. FOr instance, I am not sure exactly how the numbers are determined in KSP required for orbiting each body. Kerbin is at 70x70 km orbit a 2.4 km/s orbit(?) With Earth, if I am correct, there is a minimum orbital velocity of ~7.8 km in a 100x100 km orbit.

I've been doing some research on this, and so far I found a few equations. The simplest one I found, (this is assuming the orbit is near circular) is

Velocity = Gravitation constant * (Mass 1 and Mass 2) over distance between the two objects' centres of mass. I understand all the variables required except for G. (Or you can do the first part and do it as such: if the orbiting object's mass is negligible, you simply use the mass of the orbited body instead.)

G= 6.673 84(80) * 10^-10 m^3 kg^-1 s^1 = 6.673 84(80) * 10^-11 N m^2 kg^-2

I'm not exactly sure what I am looking at with that, nor what those variables are for. That whole thing is extremely confusing. Explanation, please?

Also, do you need to know the surface gravity of the celestial body(s) or does the equation take care of all of that for you? Or do you not need to know?

And one more thing, how does one determine the mass of a celestial body? Is there a simple, or easy equation for this or is it all a tedious long process? I've been doing research on that latter part but so far not getting any straight answers.

Here's a hypothetical scenario that I want to know, but also to set an example of what I am looking for:

Suppose there is a ship exploring a hypothetical moon in a hypothetical solar system. The moon has a gravity well of 0.49 g, and it has an equatorial radius of roughly 4,368.9 km. The ship would orbit around this moon at a semi major axis of 4,645.9 km. The orbit is assumed to be nearly circular and we're ignoring atmospheric drag of any kind.

With those numbers, if that is enough, can we determine the orbital velocity of the ship around that moon at that height? And can we determine the mass of this moon, and maybe density? What math is done here?

[AbacusWizard]

Good questions! Here's the basics:

First, gravity is an attractive force between any two objects with mass; its strength is proportional to the masses of the two objects and inversely proportional to the (squared) distance between them. So far this would suggest

Fgrav = M•m / r^2

where M and m are the two masses and r is the distance between them. However, the units of measurement don't work out, so another multiplier is needed. It turns out that that multiplier is the same no matter what objects we're dealing with, so we call it a universal constant: specifically, the Gravitational Constant or Newton's Constant, G. This gives us

Fgrav = G•M•m / r^2

Meanwhile: if no force is acting on a moving object, it will continue to travel at constant speed in a straight line. If an object is to move along a curved path, a force must constantly act upon it to cause the curvature. In the case of circular motion at constant speed, the required force is proportional to mass (more massive object means more force is needed) and speed (faster speed means more force is needed), and inversely proportional to the radius of the circle (larger circle means gentler curve, so less force is needed). Also, the speed has to be squared to get the units to work out right. This is called a "centripetal" force, because it needs to be directed towards the center of the circle (perpendicular to the object's velocity).

Fcent = m•v^2 / r

In the case of an orbiting object, the centripetal force is provided by gravity, so we can set them equal to each other:

G•M•m / r^2 = m•v^2 / r

where M is the mass of, let's say, the planet and m is the mass of the spaceship orbiting it at a distance r from the planet's center.

The variable m shows up on both sides, so we can divide it out, getting

G•M / r^2 = v^2 / r

This is important because it means the mass of the spaceship doesn't matter: two spaceships of different masses can orbit together!

We can also multiply both sides by r, canceling out the r on the right side and one of the r's on the left side:

G•M / r = v^2

If we now take the square root, we have a formula for orbital velocity:

v = sqrt (G•M/r)

In other words, if you want to orbit a planet with mass M at a constant distance r from the center of the planet, you must be moving at that speed. Note that things like the radius of the planet, its density, its surface gravity, the ship's mass, etc don't really matter; all you need to know is the mass of the planet and the radius of your desired orbit.

Of course, the game doesn't tell you r; it tells you your altitude--how far you are from the planet's surface (let's call that h for height). To get r you'll need to add the planet's radius, R, to h; r=R+h. If we substitute that into the formula we get

v = sqrt ( G M / (R+h) )

So! Look up the planet's mass and radius (on the map screen, focus on the planet and click the info button), look up the universal gravitational constant, decide what height you want to orbit at (make sure it's above the atmosphere if there is one or the highest mountain if there isn't), plug all those numbers into the formula, and you'll have the required orbital velocity.

[Steel]

I understand all the variables required except for G

Also just on top of what AbacusWizard said, G in this equation isn't a variable, but the gravitational constant, so there's no need to worry about it, you just plug in the number into your calculations every time a G comes up.

The moon has a gravity well of 0.49 g

Can you define a little better what this means? Acceleration due to gravity (which is what is usually measured in g's) varies with distance from the body, so you need to define where you have an acceleration of 0.49 g to have this make any sence.

[RocketPropelledGiraffe]

To expand on AbacusWizard's calculations, which cover your first question very well, you can determine the mass M of your moon from the given gravitational data. As Steel said you need to know at what altitude the acceleration a = 0.49 g are measured.

Newton said F=m*a, the gravitation tells you F=G*M*m/r^2 => m*a=G*M*m/r^2 =>M=a*r^2/G

Put in the correct distance r for your given accelaration (e.g. at sea level 4,368.9 km), and you get the mass M.

[Red Iron Crown]

First, lets find the mass of the planet using Newton's Law of Universal Gravitation, assuming that 4.80m/s (0.49G) is surface gravity (we can ignore the ship mass by assuming it is 1kg, so F=4.8N):

F = Gm/r² rearranges to:

m = Fr²/G

m = (4.8)*(4368900)²/(6.67e-11)

m = 1.37e24 kg

Now we can apply the Vis Viva equation to determine speed (we'll assume circular, so a = r):

v² = GM((2/r)-(1/a))

v = sqrt(GM((2/r)-(1/a)))

v = sqrt(6.67e-11*1.37e24*((2/4645900)-(1/4645900)))

v = 4434.9m/s

If the orbit is non-circular, you can substitute any altitude (from planet center) for r and calculate the speed for that altitude.

Edited June 1, 2015 by Red Iron Crown
Typo

[Der Anfang]

First, lets find the mass of the planet using Newton's Law of Universal Gravitation, assuming that 4.80m/s (0.49G) is surface gravity (we can ignore the ship mass by assuming it is 1kg, so F=4.8N):

F = Gm/r² rearranges to:

m = Fr²/G

m = (4.8)*(4368900)²/(6.67e-11)

m = 1.37e24 kg

Now we can apply the Vis Viva equation to determine speed (we'll assume circular, so a = r):

v² = GM((2/r)-(1/a))

v = sqrt(GM((2/r)-(1/a)))

v = sqrt(6.67e-11*1.37e24*((2/4645900)-(1/4645900)))

v = 4434.9m/s

If the orbit is non-circular, you can substitute any altitude (from planet center) for r and calculate the speed for that altitude.

Hello! I like your response and this is extremely helpful. However, I have run into a big problem using this information that you have provided that I can't seem to resolve to help me confirm this answer (yet, this is the best answer I have gotten in regard to calculating a celestial body's mass). I will show you my work and my answer. With that, I hope you can help me figure out what I did wrong and what I need to do?

I tried to use this equation to determine the mass of the Earth, to help test the accuracy of this answer you provided. We already know that the Earth's mass is 5.97219×10²⁴ kg. But my answer led me way off course (presumably).

The numbers I used:

1 m/s^2 = 0.1019 g (earth gravity)

1N = kg * m/2^2. Thus, 1 Earth gravity = (surface grav) 9.8 m/s^2 = 9.8N(is this N correct?)

1 R⊕ ≈ 6,378.1 km.

1 M⊕ = 5.972*10^24 kg

G = 6.673*10^-11

My work:

M = Fr^2 / G

M = (9.8) (6,378.1)^2 = 398665564.2 = 3.98*10^8

M = 3.98*10^8 / 6.673*10^-11

M = 5.96*10^18 kg

I tried to redo my calculations several times to double check and I kept getting the same end result. My result was 5.96*10^18 kg whereas the Earth's mass is supposed to be 6.972*10^24 kg. What did I do wrong? What am I missing? Either there was a miscalculation (which is most likely), or I am missing some other element, or both.

And to the others that have asked about it, I meant the surface gravity at sea level.

Edited June 8, 2015 by Der Anfang

[Windspren]

I'm trying to understand orbital velocity here. FOr instance, I am not sure exactly how the numbers are determined in KSP required for orbiting each body. Kerbin is at 70x70 km orbit a 2.4 km/s orbit(?) With Earth, if I am correct, there is a minimum orbital velocity of ~7.8 km in a 100x100 km orbit.

I've been doing some research on this, and so far I found a few equations. The simplest one I found, (this is assuming the orbit is near circular) is

Velocity = Gravitation constant * (Mass 1 and Mass 2) over distance between the two objects' centres of mass. I understand all the variables required except for G. (Or you can do the first part and do it as such: if the orbiting object's mass is negligible, you simply use the mass of the orbited body instead.)

G= 6.673 84(80) * 10^-10 m^3 kg^-1 s^1 = 6.673 84(80) * 10^-11 N m^2 kg^-2

I'm not exactly sure what I am looking at with that, nor what those variables are for. That whole thing is extremely confusing. Explanation, please?

Also, do you need to know the surface gravity of the celestial body(s) or does the equation take care of all of that for you? Or do you not need to know?

And one more thing, how does one determine the mass of a celestial body? Is there a simple, or easy equation for this or is it all a tedious long process? I've been doing research on that latter part but so far not getting any straight answers.

Here's a hypothetical scenario that I want to know, but also to set an example of what I am looking for:

Suppose there is a ship exploring a hypothetical moon in a hypothetical solar system. The moon has a gravity well of 0.49 g, and it has an equatorial radius of roughly 4,368.9 km. The ship would orbit around this moon at a semi major axis of 4,645.9 km. The orbit is assumed to be nearly circular and we're ignoring atmospheric drag of any kind.

With those numbers, if that is enough, can we determine the orbital velocity of the ship around that moon at that height? And can we determine the mass of this moon, and maybe density? What math is done here?

Good questions! Here's the basics:

First, gravity is an attractive force between any two objects with mass; its strength is proportional to the masses of the two objects and inversely proportional to the (squared) distance between them. So far this would suggest

Fgrav = M•m / r^2

where M and m are the two masses and r is the distance between them. However, the units of measurement don't work out, so another multiplier is needed. It turns out that that multiplier is the same no matter what objects we're dealing with, so we call it a universal constant: specifically, the Gravitational Constant or Newton's Constant, G. This gives us

Fgrav = G•M•m / r^2

Meanwhile: if no force is acting on a moving object, it will continue to travel at constant speed in a straight line. If an object is to move along a curved path, a force must constantly act upon it to cause the curvature. In the case of circular motion at constant speed, the required force is proportional to mass (more massive object means more force is needed) and speed (faster speed means more force is needed), and inversely proportional to the radius of the circle (larger circle means gentler curve, so less force is needed). Also, the speed has to be squared to get the units to work out right. This is called a "centripetal" force, because it needs to be directed towards the center of the circle (perpendicular to the object's velocity).

Fcent = m•v^2 / r

In the case of an orbiting object, the centripetal force is provided by gravity, so we can set them equal to each other:

G•M•m / r^2 = m•v^2 / r

where M is the mass of, let's say, the planet and m is the mass of the spaceship orbiting it at a distance r from the planet's center.

The variable m shows up on both sides, so we can divide it out, getting

G•M / r^2 = v^2 / r

This is important because it means the mass of the spaceship doesn't matter: two spaceships of different masses can orbit together!

We can also multiply both sides by r, canceling out the r on the right side and one of the r's on the left side:

G•M / r = v^2

If we now take the square root, we have a formula for orbital velocity:

v = sqrt (G•M/r)

In other words, if you want to orbit a planet with mass M at a constant distance r from the center of the planet, you must be moving at that speed. Note that things like the radius of the planet, its density, its surface gravity, the ship's mass, etc don't really matter; all you need to know is the mass of the planet and the radius of your desired orbit.

Of course, the game doesn't tell you r; it tells you your altitude--how far you are from the planet's surface (let's call that h for height). To get r you'll need to add the planet's radius, R, to h; r=R+h. If we substitute that into the formula we get

v = sqrt ( G M / (R+h) )

So! Look up the planet's mass and radius (on the map screen, focus on the planet and click the info button), look up the universal gravitational constant, decide what height you want to orbit at (make sure it's above the atmosphere if there is one or the highest mountain if there isn't), plug all those numbers into the formula, and you'll have the required orbital velocity.

[Red Iron Crown]

My work:

M = Fr^2 / G

M = (9.8) (6,378.1)^2 = 398665564.2 = 3.98*10^8

M = 3.98*10^8 / 6.673*10^-11

M = 5.96*10^18 kg

I tried to redo my calculations several times to double check and I kept getting the same end result. My result was 5.96*10^18 kg whereas the Earth's mass is supposed to be 6.972*10^24 kg. What did I do wrong? What am I missing? Either there was a miscalculation (which is most likely), or I am missing some other element, or both.

You used kilometers instead of meters for the radius. Let's recalculate with meters:

M = Fr^2 / G

M = (9.8) (6,378,100)^2 / 6.673e-11

M = ~5.974e24 kg

Still wrong, but at least we're at the right order of magnitude. I believe the rest can be accounted for by recognizing one of the assumptions of Newton's Law of Gravitation: All masses are point masses. At the distances used in most orbital calculations this isn't all that significant, but close to the surface a significant portion of Earth's mass is not attracting directly toward the center of the planet. I think what we've calculated above is the equivalent central point mass that would have the same attraction as the larger, more distributed mass of the real Earth.

I ignored this effect in my earlier reply because I thought we were talking about KSP physics, where planets are modeled as point masses at the center of a massless crust.

[cybersol]

We already know that the Earth's mass is...

My work:

M = Fr^2 / G

M = (9.8) (6,378.1)^2 = 398665564.2 = 3.98*10^8

M = 3.98*10^8 / 6.673*10^-11

M = 5.96*10^18 kg

You want to use 6378100 for r to get the units to cancel. I recommend you write down all the units for every value and cancel them explicitly (meters over meters cancel, meters * meters is meters^2). See this site about the proper way to cancel units.

EDIT: 10 minute Ninja

Edited June 8, 2015 by cybersol

[Der Anfang]

You used kilometers instead of meters for the radius. Let's recalculate with meters:

M = Fr^2 / G

M = (9.8) (6,378,100)^2 / 6.673e-11

M = ~5.974e24 kg

Still wrong, but at least we're at the right order of magnitude. I believe the rest can be accounted for by recognizing one of the assumptions of Newton's Law of Gravitation: All masses are point masses. At the distances used in most orbital calculations this isn't all that significant, but close to the surface a significant portion of Earth's mass is not attracting directly toward the center of the planet. I think what we've calculated above is the equivalent central point mass that would have the same attraction as the larger, more distributed mass of the real Earth.

I ignored this effect in my earlier reply because I thought we were talking about KSP physics, where planets are modeled as point masses at the center of a massless crust.

Ah okay. Also, it's actually pretty close, if not spot on. I made a typo in regards to the Earth's mass. The first digit is supposed to be 5, but I accidentally wrote it as a 6. But neverthless, I still made the mistake of usig km instead of metres. Thanks again! This clears everything up! Extremely helpful.

[rcp27]

As with most things in KSP, there's a Scott Manley video for this. Have a search for "orbital mechanics on paper". He doesn't derive the basic equations from scratch, but gives the numbers and equations. Useful as a bit of an introduction.

[mikegarrison]

Once you've learned all that, the next step is to learn that it's wrong. (Usefully wrong, but wrong.) http://en.wikipedia.org/wiki/General_relativity

[Snark]

I believe the rest can be accounted for by recognizing one of the assumptions of Newton's Law of Gravitation: All masses are point masses. At the distances used in most orbital calculations this isn't all that significant, but close to the surface a significant portion of Earth's mass is not attracting directly toward the center of the planet. I think what we've calculated above is the equivalent central point mass that would have the same attraction as the larger, more distributed mass of the real Earth.

I ignored this effect in my earlier reply because I thought we were talking about KSP physics, where planets are modeled as point masses at the center of a massless crust.

Actually, this doesn't matter and isn't a source of error. Modeling a spherical planet as a point mass is correct and accurate, and isn't just a convenient approximation; Newton's law of gravitation makes no assumption about point masses.

The reason this works is math: it turns out that the gravitational field of a spherically symmetrical mass distribution is exactly, mathematically identical to that of a point mass at its center. (As long as you're outside the sphere.) It's straightforward to prove if you solve the integral for the spherical mass distribution.

Of course, a real planet isn't perfectly spherical, there are asymmetries in the gravitational field due to mountains, oceans, varying crust thickness, etc., and KSP does ignore that. But that's a different issue entirely.

[mikegarrison]

Actually, this doesn't matter and isn't a source of error. Modeling a spherical planet as a point mass is correct and accurate, and isn't just a convenient approximation; Newton's law of gravitation makes no assumption about point masses.

The reason this works is math: it turns out that the gravitational field of a spherically symmetrical mass distribution is exactly, mathematically identical to that of a point mass at its center. (As long as you're outside the sphere.) It's straightforward to prove if you solve the integral for the spherical mass distribution.

It's dead simple to show this with calculus. But one of the more interesting things I learned in college is that Issac Newton first figured this out using calculus, but then spent an inordinate amount of time resolving it using only algebra and geometry. It's a much harder problem to do it that way. But Newton was keeping his invention of calculus to himself as his own private tool.

[Red Iron Crown]

Actually, this doesn't matter and isn't a source of error. Modeling a spherical planet as a point mass is correct and accurate, and isn't just a convenient approximation; Newton's law of gravitation makes no assumption about point masses.

The reason this works is math: it turns out that the gravitational field of a spherically symmetrical mass distribution is exactly, mathematically identical to that of a point mass at its center. (As long as you're outside the sphere.) It's straightforward to prove if you solve the integral for the spherical mass distribution.

Of course, a real planet isn't perfectly spherical, there are asymmetries in the gravitational field due to mountains, oceans, varying crust thickness, etc., and KSP does ignore that. But that's a different issue entirely.

There's a reason I'm not a physicist. Thanks for the correction.

[AbacusWizard]

It's straightforward to prove if you solve the integral for the spherical mass distribution.

It's even more straightforward to prove--no integrals required!--if you use a gravitational version of Gauss's Law. Replace electric flux with gravitational flux, replace charge with mass, and the equations should work out exactly the same. If the region you're investigating contains (uniformly distributed) mass, it can be treated as located at a single point; if the region you're investigating contains no mass, the gravitational flux should be zero. (Neat result: in the interior of a hollow planet, there is no gravity!)

[Snark]

If the region you're investigating contains (uniformly distributed) mass, it can be treated as located at a single point

Doesn't actually have to be uniformly distributed (in the sense of "constant density throughout the volume")-- only needs to be spherically symmetrical for the point-mass equivalence to hold. For example, real planets are far from having uniformly-distributed mass-- the core is typically several times denser than the crust. But they tend to be spherically symmetrical (i.e. density is a function of radius from center, not lat/long).