So, I think I came up with a solution. No conic sections involved; the only thing I used was physics.

A few things about Kerbal Space Program that make things easier:

1. The physics engine runs off of patched conic sections. This means that when you are in orbit around an object, there are no other influences on you. Once you're in the sphere of influence, all other objects are nil.

2. This means that all planetary orbits are immutable. No orbiting around a barycenter; even a large moon will orbit around the center of its planet.

3. Objects are all assumed to be perfectly spherical for physics purposes. Of course, for landing purposes, there are cliffs and craters.

Anyway, here's what I have so far. Let me know if my math is wrong.

Kerbal Space Program has a different solar system than ours. Their sun is called Kerbol, and their planet is called Kerbin. Orbiting around Kerbin is the Mun.

Kerbin has the following relevant characteristics:

Radius: 600km
Mass: 5.2915793*10^22 kg
Its atmosphere is pretty much dissipated at 60km and totally gone at 100km.

As you can see, the place is similar in size to Earth, albeit much more dense. Going off of the formula GM / r^2, you can see that the gravitational acceleration at Kerbin's surface is 9.81m/s^2.

The Mun has the following relevant characteristics:

Radius: 200km
Mass: 9.7600236*10^20kg

The Mun has been deliberately made as easy as possible to reach. It has the following orbital characteristics:

Semi-major Axis: 12000km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A (It doesn't even have a periapsis)

We'll make the spacecraft's orbit as easy as possible as well:

Semi-major Axis: 700km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A

Let's look at the spacecraft. Its potential energy is the following:

Ep = -GMm / r
Ep = -5045142.857m, when m is the mass of the spacecraft.
In case you did the calculation yourself, the gravitational parameter of Kerbin is a little off; it's 3.5316 * 10^12 rather than 3.531515 * 10^12.

The kinetic energy is a little bit more complicated:

a = GM / r^2 (we need acceleration, not force)
a = 3.5316*10^12 / 700000^2
a = 7.20735m/s^2

Since it's a perfectly circular orbit, we can use the following equation for velocity:

a = v^2 / r

7.20735 = v^2 / 700000
5045142.857 = v^2
v = 2246.140 m/s

Using the following formula, we can get the kinetic energy:

Ek = 1/2mv^2
Ek = 1/2m * 5045142.856
Ek = 2522571.428m

I didn't know this, but apparently this is how you show that the kinetic energy of a circular orbit is always -1/2 of its potential energy. Interesting.

Now, let's make it so that the spacecraft's orbit goes all the way to the Mun. First, we have to calculate the eccentricity of this orbit:

e = (apoapsis - periapsis) / (apoapsis + periapsis)
e = (12000km - 700km) / (12000km + 700km)
e = .88976

Really elliptical, but still perfectly fine.

Next, we get the potential energy at its new apoapsis:

Ep = -GM * mass / r
Ep = -294300m

I wasn't sure what to do from here, so I looked up some more formulas. Apparently, this formula is the total energy of an elliptical orbit:

Ep = 2 * Etotal / (1 + e)
-294300m = 2 * Etotal / 1.88976So, I think I came up with a solution. No conic sections involved; the only thing I used was physics.

A few things about Kerbal Space Program that make things easier:

1. The physics engine runs off of patched conic sections. This means that when you are in orbit around an object, there are no other influences on you. Once you're in the sphere of influence, all other objects are nil.

2. This means that all planetary orbits are immutable. No orbiting around a barycenter; even a large moon will orbit around the center of its planet.

3. Objects are all assumed to be perfectly spherical for physics purposes. Of course, for landing purposes, there are cliffs and craters.

Anyway, here's what I have so far. Let me know if my math is wrong.

Kerbal Space Program has a different solar system than ours. Their sun is called Kerbol, and their planet is called Kerbin. Orbiting around Kerbin is the Mun.

Kerbin has the following relevant characteristics:

Radius: 600km
Mass: 5.2915793*10^22 kg
Its atmosphere is pretty much dissipated at 60km and totally gone at 100km.

As you can see, the place is similar in size to Earth, albeit much more dense. Going off of the formula GM / r^2, you can see that the gravitational acceleration at Kerbin's surface is 9.81m/s^2.

The Mun has the following relevant characteristics:

Radius: 200km
Mass: 9.7600236*10^20kg

The Mun has been deliberately made as easy as possible to reach. It has the following orbital characteristics:

Semi-major Axis: 12000km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A (It doesn't even have a periapsis)

We'll make the spacecraft's orbit as easy as possible as well:

Semi-major Axis: 700km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A

Let's look at the spacecraft. Its potential energy is the following:

Ep = -GMm / r
Ep = -5045142.857m, when m is the mass of the spacecraft.
In case you did the calculation yourself, the gravitational parameter of Kerbin is a little off; it's 3.5316 * 10^12 rather than 3.531515 * 10^12.

The kinetic energy is a little bit more complicated:

a = GM / r^2 (we need acceleration, not force)
a = 3.5316*10^12 / 700000^2
a = 7.20735m/s^2

Since it's a perfectly circular orbit, we can use the following equation for velocity:

a = v^2 / r

7.20735 = v^2 / 700000
5045142.857 = v^2
v = 2246.140 m/s

Using the following formula, we can get the kinetic energy:

Ek = 1/2mv^2
Ek = 1/2m * 5045142.856
Ek = 2522571.428m

I didn't know this, but apparently this is how you show that the kinetic energy of a circular orbit is always -1/2 of its potential energy. Interesting.

Now, let's make it so that the spacecraft's orbit goes all the way to the Mun. First, we have to calculate the eccentricity of this orbit:

e = (apoapsis - periapsis) / (apoapsis + periapsis)
e = (12000km - 700km) / (12000km + 700km)
e = .88976

Really elliptical, but still perfectly fine.

Next, we get the potential energy at its new apoapsis:

Ep = -GM * mass / r
Ep = -294300m

I wasn't sure what to do from here, so I looked up some more formulas. Apparently, this formula is the total energy of an elliptical orbit:

Ep = 2 * Etotal / (1 + e)
-294300m = 2 * Etotal / 1.88976
Etotal = -278078.184m

From here, thanks to good ol' conservation of momentum, we can figure out how fast the rocket needs to be going at its periapsis:

Etotal = Ep + Ek
-278078.184m = -5045142.857m + Ek
Ek = 4767061.673m

Ek = 1/2mv^2So, I think I came up with a solution. No conic sections involved; the only thing I used was physics.

A few things about Kerbal Space Program that make things easier:

1. The physics engine runs off of patched conic sections. This means that when you are in orbit around an object, there are no other influences on you. Once you're in the sphere of influence, all other objects are nil.

2. This means that all planetary orbits are immutable. No orbiting around a barycenter; even a large moon will orbit around the center of its planet.

3. Objects are all assumed to be perfectly spherical for physics purposes. Of course, for landing purposes, there are cliffs and craters.

Anyway, here's what I have so far. Let me know if my math is wrong.

Kerbal Space Program has a different solar system than ours. Their sun is called Kerbol, and their planet is called Kerbin. Orbiting around Kerbin is the Mun.

Kerbin has the following relevant characteristics:

Radius: 600km
Mass: 5.2915793*10^22 kg
Its atmosphere is pretty much dissipated at 60km and totally gone at 100km.

As you can see, the place is similar in size to Earth, albeit much more dense. Going off of the formula GM / r^2, you can see that the gravitational acceleration at Kerbin's surface is 9.81m/s^2.

The Mun has the following relevant characteristics:

Radius: 200km
Mass: 9.7600236*10^20kg

The Mun has been deliberately made as easy as possible to reach. It has the following orbital characteristics:

Semi-major Axis: 12000km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A (It doesn't even have a periapsis)

We'll make the spacecraft's orbit as easy as possible as well:

Semi-major Axis: 700km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A

Let's look at the spacecraft. Its potential energy is the following:

Ep = -GMm / r
Ep = -5045142.857m, when m is the mass of the spacecraft.
In case you did the calculation yourself, the gravitational parameter of Kerbin is a little off; it's 3.5316 * 10^12 rather than 3.531515 * 10^12.

The kinetic energy is a little bit more complicated:

a = GM / r^2 (we need acceleration, not force)
a = 3.5316*10^12 / 700000^2So, I think I came up with a solution. No conic sections involved; the only thing I used was physics.

A few things about Kerbal Space Program that make things easier:

1. The physics engine runs off of patched conic sections. This means that when you are in orbit around an object, there are no other influences on you. Once you're in the sphere of influence, all other objects are nil.

2. This means that all planetary orbits are immutable. No orbiting around a barycenter; even a large moon will orbit around the center of its planet.

3. Objects are all assumed to be perfectly spherical for physics purposes. Of course, for landing purposes, there are cliffs and craters.

Anyway, here's what I have so far. Let me know if my math is wrong.

Kerbal Space Program has a different solar system than ours. Their sun is called Kerbol, and their planet is called Kerbin. Orbiting around Kerbin is the Mun.

Kerbin has the following relevant characteristics:

Radius: 600km
Mass: 5.2915793*10^22 kg
Its atmosphere is pretty much dissipated at 60km and totally gone at 100km.

As you can see, the place is similar in size to Earth, albeit much more dense. Going off of the formula GM / r^2, you can see that the gravitational acceleration at Kerbin's surface is 9.81m/s^2.

The Mun has the following relevant characteristics:

Radius: 200km
Mass: 9.7600236*10^20kg

The Mun has been deliberately made as easy as possible to reach. It has the following orbital characteristics:

Semi-major Axis: 12000km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A (It doesn't even have a periapsis)

We'll make the spacecraft's orbit as easy as possible as well:

Semi-major Axis: 700km
Eccentricity: 0
Inclination: 0
Ascending Node: N/A
Argument of Periapsis: N/A

Let's look at the spacecraft. Its potential energy is the following:

Ep = -GMm / r
Ep = -5045142.857m, when m is the mass of the spacecraft.
In case you did the calculation yourself, the gravitational parameter of Kerbin is a little off; it's 3.5316 * 10^12 rather than 3.531515 * 10^12.

The kinetic energy is a little bit more complicated:

a = GM / r^2 (we need acceleration, not force)
a = 3.5316*10^12 / 700000^2
a = 7.20735m/s^2

Since it's a perfectly circular orbit, we can use the following equation for velocity:

a = v^2 / r

7.20735 = v^2 / 700000
5045142.857 = v^2
v = 2246.140 m/s

Using the following formula, we can get the kinetic energy:

Ek = 1/2mv^2
Ek = 1/2m * 5045142.856
Ek = 2522571.428m

I didn't know this, but apparently this is how you show that the kinetic energy of a circular orbit is always -1/2 of its potential energy. Interesting.

Now, let's make it so that the spacecraft's orbit goes all the way to the Mun. First, we have to calculate the eccentricity of this orbit:

e = (apoapsis - periapsis) / (apoapsis + periapsis)
e = (12000km - 700km) / (12000km + 700km)
e = .88976

Really elliptical, but still perfectly fine.

Next, we get the potential energy at its new apoapsis:

Ep = -GM * mass / r
Ep = -294300m

I wasn't sure what to do from here, so I looked up some more formulas. Apparently, this formula is the total energy of an elliptical orbit:

Ep = 2 * Etotal / (1 + e)
-294300m = 2 * Etotal / 1.88976
Etotal = -278078.184m

From here, thanks to good ol' conservation of momentum, we can figure out how fast the rocket needs to be going at its periapsis:

Etotal = Ep + Ek
-278078.184m = -5045142.857m + Ek
Ek = 4767061.673m

Ek = 1/2mv^2
4767061.673m = 1/2mv^2
v^2 = 9534123.346
v = 3087.738m/s

Seeing as how the escape velocity is 3431.03m/s, (Set the eccentricity to 1) this sounds about right.

Now that we have the change in velocity required to intersect the Mun's orbit, we now have to figure out how to make sure that the Mun will be there when we arrive. To do that, we need the period of our new orbit.

Some really smart German guy made the following equation for orbital period:

T = 2π√(a^3 / (GM)), when a is the semi-major axis.

We get the semi-major axis by adding the periapsis and apoapsis together and dividing by two, to get 6350000.

T = 2π√(6350000^3 / GM)
T = 53500.11 seconds

Handy.

From there, since the Mun's orbit is circular, we just find the fraction that this is of the Mun's period:

Tm = 138984 seconds
T = 53500.11 seconds

T / Tm = .19247 revolutions, or 1.209 radians, or 69.271 degrees from the apoapsis. Alternatively, it's 110.729 degrees from the spacecraft. Delta-V required is 841.598 m/s.

This is a really special case; a lot of the techniques I just used will not work for elliptical starting orbits. Any insight would be greatly appreciated.

Love,
Mike
a = 7.20735m/s^2

Since it's a perfectly circular orbit, we can use the following equation for velocity:

a = v^2 / r

7.20735 = v^2 / 700000
5045142.857 = v^2
v = 2246.140 m/s

Using the following formula, we can get the kinetic energy:

Ek = 1/2mv^2
Ek = 1/2m * 5045142.856
Ek = 2522571.428m

I didn't know this, but apparently this is how you show that the kinetic energy of a circular orbit is always -1/2 of its potential energy. Interesting.

Now, let's make it so that the spacecraft's orbit goes all the way to the Mun. First, we have to calculate the eccentricity of this orbit:

e = (apoapsis - periapsis) / (apoapsis + periapsis)
e = (12000km - 700km) / (12000km + 700km)
e = .88976

Really elliptical, but still perfectly fine.

Next, we get the potential energy at its new apoapsis:

Ep = -GM * mass / r
Ep = -294300m

I wasn't sure what to do from here, so I looked up some more formulas. Apparently, this formula is the total energy of an elliptical orbit:

Ep = 2 * Etotal / (1 + e)
-294300m = 2 * Etotal / 1.88976
Etotal = -278078.184m

From here, thanks to good ol' conservation of momentum, we can figure out how fast the rocket needs to be going at its periapsis:

Etotal = Ep + Ek
-278078.184m = -5045142.857m + Ek
Ek = 4767061.673m

Ek = 1/2mv^2
4767061.673m = 1/2mv^2
v^2 = 9534123.346
v = 3087.738m/s

Seeing as how the escape velocity is 3431.03m/s, (Set the eccentricity to 1) this sounds about right.

Now that we have the change in velocity required to intersect the Mun's orbit, we now have to figure out how to make sure that the Mun will be there when we arrive. To do that, we need the period of our new orbit.

Some really smart German guy made the following equation for orbital period:

T = 2π√(a^3 / (GM)), when a is the semi-major axis.

We get the semi-major axis by adding the periapsis and apoapsis together and dividing by two, to get 6350000.

T = 2π√(6350000^3 / GM)
T = 53500.11 seconds

Handy.

From there, since the Mun's orbit is circular, we just find the fraction that this is of the Mun's period:

Tm = 138984 seconds
T = 53500.11 seconds

T / Tm = .19247 revolutions, or 1.209 radians, or 69.271 degrees from the apoapsis. Alternatively, it's 110.729 degrees from the spacecraft. Delta-V required is 841.598 m/s.

This is a really special case; a lot of the techniques I just used will not work for elliptical starting orbits. Any insight would be greatly appreciated.

Love,
Mike
4767061.673m = 1/2mv^2
v^2 = 9534123.346
v = 3087.738m/s

Seeing as how the escape velocity is 3431.03m/s, (Set the eccentricity to 1) this sounds about right.

Now that we have the change in velocity required to intersect the Mun's orbit, we now have to figure out how to make sure that the Mun will be there when we arrive. To do that, we need the period of our new orbit.

Some really smart German guy made the following equation for orbital period:

T = 2π√(a^3 / (GM)), when a is the semi-major axis.

We get the semi-major axis by adding the periapsis and apoapsis together and dividing by two, to get 6350000.

T = 2π√(6350000^3 / GM)
T = 53500.11 seconds

Handy.

From there, since the Mun's orbit is circular, we just find the fraction that this is of the Mun's period:

Tm = 138984 seconds
T = 53500.11 seconds

T / Tm = .19247 revolutions, or 1.209 radians, or 69.271 degrees from the apoapsis. Alternatively, it's 110.729 degrees from the spacecraft. Delta-V required is 841.598 m/s.

This is a really special case; a lot of the techniques I just used will not work for elliptical starting orbits. Any insight would be greatly appreciated.

Love,
Mike
Etotal = -278078.184m

From here, thanks to good ol' conservation of momentum, we can figure out how fast the rocket needs to be going at its periapsis:

Etotal = Ep + Ek
-278078.184m = -5045142.857m + Ek
Ek = 4767061.673m

Ek = 1/2mv^2
4767061.673m = 1/2mv^2
v^2 = 9534123.346
v = 3087.738m/s

Seeing as how the escape velocity is 3431.03m/s, (Set the eccentricity to 1) this sounds about right.

Now that we have the change in velocity required to intersect the Mun's orbit, we now have to figure out how to make sure that the Mun will be there when we arrive. To do that, we need the period of our new orbit.

Some really smart German guy made the following equation for orbital period:

T = 2π√(a^3 / (GM)), when a is the semi-major axis.

We get the semi-major axis by adding the periapsis and apoapsis together and dividing by two, to get 6350000.

T = 2π√(6350000^3 / GM)
T = 53500.11 seconds

Handy.

From there, since the Mun's orbit is circular, we just find the fraction that this is of the Mun's period:

Tm = 138984 seconds
T = 53500.11 seconds

T / Tm = .19247 revolutions, or 1.209 radians, or 69.271 degrees from the apoapsis. Alternatively, it's 110.729 degrees from the spacecraft. Delta-V required is 841.598 m/s.

This is a really special case; a lot of the techniques I just used will not work for elliptical starting orbits. Any insight would be greatly appreciated.

Love,
Mike