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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 ...


11

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

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 ...


5

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

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 ...


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

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

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 ...


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

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

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

How can Earth-Centered Inertial (ECI) coordinates be inertial if Earth's orbital motion is always accelerating? It is true that "Earth-Centered Inertial" is a bit of a misnomer. What this means is that one has to account for the fictitious acceleration that results from the acceleration of the frame of reference. Unlike the fictitious accelerations that ...


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; $...


2

All right, I'd approach this in the following manner. Part 1: True Anomaly $\theta$ at $r$ in terms of Elliptical Orbit Eccentricity $e$ The first place I'd start is the Vis-Viva Equation, which, for all orbits around a particular body with a particular gravitational parameter $\mu$ links relative speed $v$ with radial distance $r$ and semimajor-axis $a$. ...


2

In his excellent answer Notovy used the vis viva equation to demonstrate same speeds implies r = a. A basic property points on an ellipse: sum of distance from one focus plus distance to the other focus is 2a. So if the radius vector has the same length as semi major axis a, that implies the end of the radius vector lies on the end of the ellipse's semi-...


2

Earth's sphere of influence has a radius of about 924000km. A highly eccentric orbit with a perigee at 100km altitude and an apogee at the SOI radius has a semimajor axis of 465239km. Throwing that into the vis-viva equation $v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$ where $G$ is the gravitational constant, $M$ the mass of earth, $r$ the orbital ...


2

Energy is always conserved, but different oberservers will disagree about how much energy there is, and what forms it takes. Also you have to be sure to include the whole system. Let's return to the Phobos example from the linked question, but be a bit more careful. Suppose what we actually do is use our teraton nukes to split Phobos into two equal halves ...


2

There are no limits on the direction. The Vis-viva equation will give you a speed. Assuming point-masses and sticking to classical mechanics, the Vis-Viva equation does not care at all about what direction you point your velocity in; It's merely an equation based on how the total Orbital Energy (which is the same for all orbits with the same semimajor ...


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 ...


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