Kamis, 16 Juli 2015

Ied Mubarak All!

Well, it's 20.54 at night, and 16/07/2015
And, it's the last day of ramadhan!
*uraaa
Well, tomorow, there will be many foods here, in my house, and when i'm bored with the food here
I can go to my neighbour or other family member house, and eat!
And, there is a possibility to be a rich person tomorow

Ok, it's a little bit loud right here, and when i ask one of my friend, she said, it's loud too there
the sounds of firework, firecracker, and other stuff (Including takbir) are here, i'm afraid that one of of that sounds around me is a military class explosion (?)
Well, if you want some kind of artilery battle feeling, you can wait till' the new year, and come to the nearest firework party, but , dont get too close, get a little bit far (500m maybe work), and look for a quiet place, wait till 00.00, and feel it

Ok, i'm talking too much shit now
I need to continue my kerbal space program rocket
Well, if you think, kerbal space program is only an ordinary game, i'll proof you are wrong
I will tell you how to orbit

First, you need to adjust your orbiting speed
The relation between orbital speed and acceleration is given by the formula:
a={\frac {v^{2}}{r}},
where a is the acceleration due to gravity, v is the horizontal speed, and r is the radius of orbit.
Of course, gravity varies depending on your distance from the planet, so we also need the following formula to determine a based on your altitude:
a=g\left({\frac {R}{R+h}}\right)^{2},
where g is the acceleration due to gravity at sea level (9.807 m/s2), R is the radius of Kerbin (600 km), and h is the altitude of your orbit.
Note: Since the radius of your orbit r is equal to the radius of Kerbin R plus your altitude h, we can substitute R + h for r.
Now we have two expressions for a, so we set them on opposite sides of an equation and simplify:
{\begin{aligned}g\left({\frac {R}{(R+h)}}\right)^{2}&={\frac {v^{2}}{r}}\\g\left({\frac {R}{(R+h)}}\right)^{2}&={\frac {v^{2}}{R+h}}\\g{\frac {R^{2}}{(R+h)^{2}}}&={\frac {v^{2}}{R+h}}\\g{\frac {R^{2}}{R+h}}&=v^{2}\\v^{2}&=g{\frac {R^{2}}{R+h}}\\v&={\sqrt {g{\frac {R^{2}}{R+h}}}}\\v&=R{\sqrt {{\frac {g}{R+h}}}}\\\end{aligned}}
Finally, substituting known values for g and R,
v=600\ 000\ {\mathrm {m}}{\sqrt {{\frac {9.807\ {\mathrm {m}}/{\mathrm {s}}^{2}}{600\ 000\ {\mathrm {m}}+h}}}}

And then, the orbital period
From the basic mechanics formula:
d=vt
We know v from the above, and d is simply the circumference of a circle with a radius equal to your orbital altitude plus the radius of Kerbin:
t=2\pi {\frac {600\ 000\ {\mathrm {m}}+h}{v}}

And, last but not least, orbital equation
For this section it is assumed that the craft in question is orbiting a particular body and is a lot smaller than the body that it is orbiting. Other then in the general orbital equation and the energy equation, the orbits are assumed to be bounded orbits completely within the body's sphere of influence and no others. This section also assumes that classical physics applies in the game. All distances and radii are from the center of the body being orbited and the sea level radius of the body needs to be added to the altitude given in game to recover the values here. The section uses the standard gravitational parameter \mu =GM which G is Newton's Gravitational Constant and M is the mass of the body that is being orbited. From here on out m will refer to the mass of the craft.
When the craft is orbiting a body, the distance between the craft and the body is given by the equation
r(\phi )={\frac {l^{2}}{\mu m^{2}}}{\frac {1}{1+\epsilon cos\phi }}
where l is the angular momentum of the craft, \epsilon is the eccentricity of the orbit, and \phi is an angle parameter for the orbit. \phi runs from 0\deg to 360\deg , l=m\cdot r\cdot v at apoapsis or periapsis, and \epsilon \geq 0. When \epsilon <1 the orbit is bounded and is an ellipse and when \epsilon \geq 1 the orbit is unbounded. At \phi =0\deg the craft is at the periapsis and at \phi =180\deg the craft is at the apoapsis. At periapsis,
r_{p}={\frac {l^{2}}{\mu m^{2}}}{\frac {1}{1+\epsilon }}
and at apoapsis,
r_{a}={\frac {l^{2}}{\mu m^{2}}}{\frac {1}{1-\epsilon }}.
The semimajor axis is
a={\frac {l^{2}}{\mu m^{2}}}{\frac {1}{1-\epsilon ^{2}}}
and the semiminor axis is
b={\frac {l^{2}}{\mu m^{2}}}{\frac {1}{{\sqrt {1-\epsilon ^{2}}}}}.
\epsilon ={\frac {{\sqrt {a^{2}-b^{2}}}}{a}}
Given the results for r_{p} and r_{a} above,
l^{2}=2\mu m^{2}{\frac {r_{p}r_{a}}{r_{p}+r_{a}}} and
\epsilon ={\frac {r_{a}-r_{p}}{r_{p}+r_{a}}} thus
r(\phi )=2{\frac {r_{p}r_{a}}{r_{p}+r_{a}+(r_{a}-r_{p})cos\phi }}
The velocity for an object in elliptical orbit about a much larger object is
v={\sqrt {\mu ({\frac {2}{r}}-{\frac {1}{a}})}}={\sqrt {\mu {\frac {r_{p}^{2}+r_{a}^{2}+(r_{a}^{2}-r_{p}^{2})cos\phi }{r_{p}r_{a}(r_{p}+r_{a})}}}}={\sqrt {2\mu {\frac {r_{p}+r_{a}-r}{r(r_{p}+r_{a})}}}}.
The energy for an object in orbit is
E={\frac {\mu ^{2}m^{3}}{2l^{2}}}(\epsilon ^{2}-1)={\frac {\mu m(r_{p}+r_{a})}{4r_{p}r_{a}}}(\epsilon ^{2}-1)=-{\frac {\mu m}{r_{p}+r_{a}}}.
One should note that for bounded orbits, the energy is negative and for escape orbits, energy is non-negative.
The orbital period is
\tau ={\sqrt {{\frac {4\pi ^{2}}{\mu }}a^{3}}}={\sqrt {{\frac {\pi ^{2}(r_{p}+r_{a})^{3}}{2\mu }}}}.
Let r_{{c,1}} and r_{{o,1}} be the two apsides of a given orbit and let a craft execute a burn imparting a specific \Delta v all at the single point in time the craft is at r_{{c,1}}. For this problem \Delta v can be either positive or negative. If it is positive then the thrust is to speed up the craft but if it is negative the ture delta v is the oppositie of that in the problem but the thrust is to slow down the craft. This problem assumes that the thust is in the same or opposite direction of travel and is not enough to reverse the craft. The new orbit will have apsides of r_{{c,2}} and r_{{o,2}} with r_{{c,2}}=r_{{c,1}}.
The speed of the craft before the burn is
v_{1}={\sqrt {2\mu {\frac {r_{{c,1}}+r_{{o,1}}-r_{{c,1}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,1}})}}}}={\sqrt {2\mu {\frac {r_{{o,1}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,1}})}}}}.
The speed of the craft after the burn is
v_{2}=v_{1}+\Delta v={\sqrt {2\mu {\frac {r_{{c,2}}+r_{{o,2}}-r_{{c,2}}}{r_{{c,2}}(r_{{c,2}}+r_{{c,2}})}}}}={\sqrt {2\mu {\frac {r_{{o,2}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,2}})}}}}={\sqrt {2\mu {\frac {r_{{o,1}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,1}})}}}}+\Delta v
which yields
\Delta v={\sqrt {{\frac {2\mu }{r_{{c,1}}}}}}({\sqrt {{\frac {r_{{o,2}}}{r_{{c,1}}+r_{{o,2}}}}}}-{\sqrt {{\frac {r_{{o,1}}}{r_{{c,1}}+r_{{o,1}}}}}})
and
r_{{o,2}}=r_{{c,1}}{\frac {r_{{c,1}}({\sqrt {2\mu {\frac {r_{{o,1}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,1}})}}}}+\Delta v)^{2}}{2\mu -r_{{c,1}}({\sqrt {2\mu {\frac {r_{{o,1}}}{r_{{c,1}}(r_{{c,1}}+r_{{o,1}})}}}}+\Delta v)^{2}}}=r_{{c,1}}{\frac {r_{{c,1}}v_{2}^{2}}{2\mu -r_{{c,1}}v_{2}^{2}}}

 Well, still thinking that game give you nothing?
Think again

Well, i want to go now
continue build some rocket
Bye!
xoxo (well, is there anyone know why "xoxo" means hug and kiss?)

Ps. Again, Ied Mubarak All!
Diposting oleh di

Tidak ada komentar:

Posting Komentar

Langganan: Posting Komentar (Atom)