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Nonrelativistic solution The variables used will be $x$ for the distance travelled $v$ for velocity $a$ for acceleration ($1~\mathrm{g}$) $t$ for the time $c$ for the speed of light. Non braking Assuming the velocity you arrive at does not matter we take the equation $$x = \frac12 a t^2\ .$$ Solve for $t$: $$t = \sqrt{\frac{2x}{a}}\ .$$ (Let’s ...


10

The eccentricity is 1.0. The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as: $$e=\frac{r_a-r_p}{r_a+r_p}$$ and the semimajor axis $a$ can as well, from: $$a=\frac{r_a+r_p}{2}$$ If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower ...


9

Breaking that down: launch due East site in the Ecuadorean Andes sometime before local midnight on a July 4 when there's a new moon Launch due east. With the exception of launch sites that cannot launch due east lest a failed launch rain debris on some other country, this is the preferable direction. Launching to the east takes advantage ...


7

This is not a recurrence for a Legendre polynomial $P_{n}(\mu)$ but an Associated Legendre Function (ALF). This1 states much more than is needed here. ALFs are often denoted $P_{n,m}(\mu)$ in geophysics. The degree $n$ and order $m$ are non-negative integers $0, 1, 2, \dots$ with $0 \le m \le n.$ Argument $\mu$ is the sine of the latitude (measured north or ...


7

The eccentricity of a radial orbit is $1$, regardless of its energy. This is a class of orbits where the type of orbit cannot be inferred from the eccentricity alone. With a "traditional" parabolic orbit of $e=1$, the angular momentum $L$ has a well defined value, but the semi-major axis $a$ is not defined. In the case of a vertical bounded free-fall orbit, ...


6

The expression on the right is meant to give the eccentricity vector but the vector notation has been lost. Here it is in this answer: $$ e = {v^2 r \over {\mu}} - {(r \cdot v ) v \over{\mu}} - {r\over{\left|r\right|}}$$ and the vector nature is not clear either. We should write it as $$ \mathbf{e} = {v^2 \mathbf{r} \over {\mu}} - {(\mathbf{r} \cdot \...


5

When you're trying to rotate something, there are two cases: 1) The torque you're applying is large compared to the angular momentum the body has, i.e. when the body isn't rotating. Then it starts to rotate in the direction of the torque you're applying. This is the more intuitive case. 2) The torque you're applying is small compared to the angular ...


5

I've looked through your numbers, and they do seem to be correct. The range values included are likely intended to be in km, but seem to be inaccurate. The range values are not official at this point in time, and do not have access to the spacecraft ephemeris data yet. I suspect when that data comes down the range values will be estimated better. Bottom ...


5

Check out the diagram at the top of the page that you got the equation from. Let's define our terms. $\dot{m}_e V_e$ is the momentum thrust term $\dot{m}_0 V_0$ is the incoming momentum term $(p_e - p_0) A_e$ is the pressure thrust term The incoming momentum term is important for jet engines because the engine swallows the incoming stream and then ...


4

Ideally if you could design the nozzle to match the exhaust pressure in a vacuum (i.e. nearly zero), the third term drops automatically. If $p_0$ is zero, then $p_e$ would have to go to zero as well because an ideally designed nozzle results in no pressure drag (i.e. ambient freestream pressure and exhaust pressure are the same). In reality, such a nozzle ...


4

Let's start by assuming you don't decelerate halfway. Work in units with $c=1$. With a constant acceleration of $a$, the rapidity $\phi=a\tau$ at a proper time $\tau$ after you start from rest, so $$\beta=\tanh a\tau,\,\gamma=\cosh a\tau,\,dx=\beta dt=\beta\gamma d\tau=\sinh a\tau d\tau,$$where $x$ is the distance travelled and $dt=\gamma d\tau$ is the ...


4

Welcome to the site! Using this tool: Observer time: 100001 years Traveler time: 22.4 years Edit: Time is fixed, I blame the google calculator


4

The potential of fly-by ping pong is pretty much unlimited, provided you have enough time to your disposal. Given an initial transfer with a perihelion slightly lower than the orbit of Venus, a Venus flyby can increase the aphelion to a bit further out than the Earth's orbit. On the following Earth flyby, the perihelion can be lowered, and you can just ...


3

I will give my current best shot at this problem, and others should feel free to strengthen the argument with additional mathematics. (Or poke holes!) You ask two questions, I will answer the first as the second has been partially answered by the update. Are there other inclination change strategies that are more efficient for some values of $\alpha$? ...


3

Given either geocentric latitude, longitude, and radius or geodetic latitude, longitude, and altitude, the computation of Earth-centered, Earth-fixed cartesian coordinates is fairly simple. For geocentric coordinates $R,\theta,\lambda$, one uses $$ \begin{aligned} R\ &\text{is the radial distance from the center of the Earth} \\ \theta\ &\text{is ...


3

I've answered some aspects of this, but considering a 1km railgun in my answer to the "parent" question. A much longer railgun doesn't make a lot of difference to the calculations there, except for the acceleration and power. The issues with reaction de-orbiting the railgun are the same. Concerning the shape of a longer railgun. Let us consider a 10g 4 km/s ...


3

Update: As of 2019-01-17, the page has been updated with range figures more in line with what you've calculated. According to the JHUAPL website (click on 'Learn more about these images ', for some reason a direct link didn't work), range should be in km: The following ancillary information is provided for each image posted: the date of the observation ...


3

note: This is a very helpful extended comment that may be of use to the OP but it can't currently be posted as a comment until this user reaches 50 reputation points. Oh this is a fantastic question. It is common to fall into the following trap when making these types of calculations. Check carefully your reference frames. Celta V numbers are all relative,...


3

Short answer: you stop nutation the way you stop any oscillation, by giving it an appropriate opposite impulse as it crosses zero, the oscillations central, zero energy point. Longer answer: The terminology in this area can be messy. So let's start with a simple model. The classic motions of a spinning top or (Navy) gyroscope are "spin", "precession", and "...


3

The article assumes that the Earth's shape is an oblate spheroid, i.e., a "squashed" sphere. If we "unsquash" everything back, including satellites' positions (which means multiplying the $z$-coordinates of the satellites by $1/\sqrt{1-\epsilon_e^2}$), the line segment connecting the new positions would intersect the sphere if and only if the line segment ...


2

Looks like they've used some figures that are very close to the sample calculations shown in the original paper with Katherine Johnson, which is available from NASA archives. The figure is for the distance of the vehicle from the center of the earth at the time of the retro rocket firing (or burnout) which initiates reentry. They've changed the original ...


2

The only equation needed to produce these curves is the thrust equation in a couple of different guises. $$F = \dot{m}v_e + (p_e-p_{atm})A_e $$ $$I_{sp}= F /(g_0 \dot{m})$$ You also need isentropic flow charts or tables to calculate the nozzle parameters that change with changing area ratio. An example isentropic flow chart is shown here, from 1953 book "...


2

This is not an answer. Here's a clever idea that doesn't work: Find the Earth's position relative to the solar system barycenter at several points in time. Find the best fit ellipse to those positions (similar to finding osculating elements, but not instantaneous ones). Look at the foci of the result to determine where the central object must be and it's ...


2

The most common method, by far, is brute force, by propagating the relevant satellite and then calculating the visibility. There are however many ways to optimize this brute force: Large delta-t until "close" and then smaller delta-t after that Using Multi-threading to reduce processing time Few github examples: https://github.com/shupp/Predict https://...


2

Dimensional analysis is a zeroth-order way to approximate (thank you Mr. Ross; 9th grade Physics). There will likely be better answers, but let's see what happens. meters/sec^2 x seconds = meters/sec 9.8 m/sec^2 x 150 sec (MECO, where you're going mostly sideways) gives ~1500 m/s delta-v People usually give something like 0.9 to 1.5 km/s when forced to ...


2

(This answer continues the one above.) A table of $P_{n,m}(\mu)$ values can be envisioned as a square array. Rows down the page are indexed by degree $n$, and columns to the right by order $m$. Due to the underlying math, the only sensible $P_{n,m}(\mu)$ values occur for $0 \le m \le n$. The effect is that the table is lower-triangular instead of square; $...


1

I’m not a space engineer, but I found this thread and I think it can answer your question. Delta V as a Function of Altitude The lift-off delta v to a $100km$ altitude is in the range of about $1.4km/s$ for an ideal system. And the answer with explained equation : To just reach a height of 150km at least once, you don't need to achieve a true orbit (...


1

The caption in the book (Sutton) reads: Normalized vehicle velocity increment as a function of normalized exhaust velocity for various payload fractions with negligible inert mass of propellant tanks. The optima of each curve are connected by a line that represents Eq. 17–9. The text reads: For a given mission, theoretically there is an optimum range ...


1

If I understand you correctly the state vector you would use would be defined in an arbitrary reference frame and you would like to find the position of the central body in that same arbitrary reference frame that would result in an orbit which includes that state vector. Mathematically there are infinitely many solutions to this problem, any position you ...


1

Is North Korea's KMS-4 in a proper Sun-synchronous orbit? tl;dr: Based on an approximate analysis of its TLE, KMS-4 is in an imperfect Sun-synchronous orbit, and will drift slightly in sun-synchrony by about 4.8 degrees each year. How do I tell? I want to understand the mechanics. tl;dr: With an inclination of about 97.4 degrees at a low altitude of ~...


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