# Tag Info

0

Have you looked at Lambert's Problem? It can be used to solve for a conic orbit that goes from pointA at timeA to pointB at timeB around the same centre of attraction. It only works for the two body problem though, so it wouldn't take into account the gravity of any other bodies. It is used in preliminary calculations for interplanetary missions.

0

I was trying to solve a similar problem, and did, with help on StackOverflow. My question was here: http://stackoverflow.com/questions/16501182/find-first-root-of-a-black-box-function-or-any-negative-value-of-same-function I asked it more abstractly. The way I saw it, you have a ship and moon (for example), and for different values of time they have a ...

2

Not only the position in the gravitational field is important, but also the velocity. Consider the Schwarzschild metric $$\text{d}\tau^2 = \left(1 - \frac{2GM}{rc^2}\right)\text{d}t^2 - \frac{1}{c^2}\left(1 - \frac{2GM}{rc^2}\right)^{-1}\left(\text{d}x^2 + \text{d}y^2 +\text{d}z^2\right),$$ where $\text{d}\tau$ is the time measured by a moving clock at ...

0

Clocks tick slower at lower altitudes. So 1. On the surface of the Earth will be the slowest. Now since the ISS has no way of knowing whether it is in orbit or in deep space, you might think that clock 2 and 3 should tick at the same rate. But instead clocks 2 and 3 will just feel like as if they were ticking at the same rate. Astronauts at 2 and 3 will not ...

1

There is a standard way to find out if the spacetime around you is curved. Surround yourself with a sphere of small test masses and wait and see what happens. If the sphere stays exactly the same shape you're in flat spacetime but if it changes shape or volume you're in a curved spacetime. In the case of the ISS the test masses nearer the Earth than you ...

3

Kepler's 3rd Law: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. If we take Jupiter's semimajor axis of 5.2 AU (avg value), then objects at 2.5 AU, 2.82 AU, 2.95 AU, 3.25 AU have orbital periods shorter than Jupiter by a factor of 3:1, 5:2, 7/3 etc. The fact that the ratios for the deepest ...

4

If you look at this problem in 2D you have the following parameters at some instant which describe your trajectory (position and velocity) around a celestial body with gravitational parameter $\mu$: radius $r$, radial velocity $\dot{r}$ and angular velocity $\omega$. There are also a few others, but these do not really matter in this problem, due to ...

3

The moon orbits around the sun, but so does the earth. They orbit together with the moon's orbit perturbed by the nearby earth. If fact, despite their different masses they experience the same acceleration, so it shouldn't be surprising that they are bound to the same orbit since they are bound to each other (i.e. at basically the same distance from the ...

-1

The answer below supports an incorrect idea. See comments. The strength of the gravitational effect on an object (e.g., a spaceship or a moon) depends both on the mass of the celestial object and the distance from that celestial object. I suspect the difficulty you're experiencing has to do with not accounting for this second effect. In essence, the Earth ...

2

I mean why should the gravity of a less massive object dominate the gravity of a more massive one? Within the Hill sphere of the Earth, objects can orbit the Earth, because in the non-rotating frame of reference centered in the Earth (moving with acceleration around the Sun, so the frame is non-inertial), the Sun's gravity force is for the most part ...

2

If I understand your question correctly you assume that the spaceship is driven by some kind of engine giving it the necessary speed to revolve around the earth. As the astronaut does not have such an engine you believe he should fall back on the earth. If this interpretation is not correct, maybe you could make your point a little more clear. There are two ...

4

Remember that for the astronaut's spaceship to be in a steady orbit, it must be moving around the earth at the appropriate velocity $v$, where $\frac{GMm}{R^2} = \frac{mv^2}{R}$ (i.e. the gravitational pull of earth is matched by the force needed to accelerate the astronaut in a circular orbit), where $M$ is the mass of the earth, $m$ of the ship, and $R$ is ...

0

The sum of two angles in different planes is known as a dogleg angle. A dogleg angle is not the same as an angle. The operation of adding two angles in different planes to get a dogleg angle is well-defined mathematically. The reasons astronomers use the the longitude of the perihelion instead of the argument of the perihelion are circular orbits and ...

1

Imagine if all the astronauts and cosmonauts inside the ISS started bouncing off the walls, would this impact the trajectory of the ISS. The physics says no. The ISS actually had a problem like this, but it does not result in orbital trajectory change. The center of mass of an object in space will move along its path regardless of motion within or about the ...

2

I'd like to add a clarification to the other answers, some of which seem to imply that the precession of Mercury's orbital perehelion is owing to general relativistic frame dragging. In particular, the statement that the Sun drags the fabric of space time around with it could be, in my opinion, misleading because most of the precession is NOT owing to "frame ...

-4

The solution of Einstein, contesting Newton´s laws, was challenged by several scientists including Dr. Thomas Van Flandern astronomer who worked at the U.S. Naval Observatory in Washington. According to them, Einstein would have gotten this information (43 "arc) and" adjusted "the arguments for the result of the equation, previously known, were achieved, ...

1

If the two objects are equal in mass (or close to it), both orbit their barycenter, which would be a point outside either body. If one object suddenly loses half its mass, the COM of the binary system moves with respect to the current locations of both objects, resulting in changes to acceleration for both ($a=\frac{GM}{ r^2}$, where r is distance to ). i.e, ...

9

Newton's original proof was in fact based on geometry (he hadn't invented calculus yet). Richard Feynman devised his own, simpler geometric proof for one of his famous lectures. You can find it in Feynman's Lost Lecture, by Goodstein & Goodstein, and in this article: Paths of the Planets from Hall & Higson. But since it's so much fun, I'll describe ...

1

The super massive black hole is creating a force acting on the material in the galaxy, but that material still has angular momentum which needs to be conserved. In a similar way, the earth is in orbit around the sun and it is bound in its orbit by the gravitational potential from the sun's mass. If you were to naively calculate the force on the earth as $F= ... 1 "Focus" is an inconvenient word if you're thinking of changing the potential, because if you do then the orbits are no longer conics and the word kind of loses its meaning. That aside, let me see if I understood your question correctly: Given a gravitational potential that's spherically symmetric around a central point$\mathbf{r}_0$, and which has a ... 0 The velocity of an orbit around some central object can be easily calculated for a circular orbit. Let us assume that there is some central Force$F=c\cdot r^\alpha$, where$c$and$\alpha$are some constants (for gravity$c=Gm_1m_2$and$\alpha=-2$). For a stable orbit, this central force must be equal to the necessary centripetal force (not balance the ... 0 The best way to explain it (and even the way Kepler's second law can be derived) is by conservation of angular momentum. The latter is given by $$\mathbf{L}=\mathbf{r}\times m\mathbf{v},$$ where$\mathbf{r}$is the position vector and$\mathbf{v}$is velocity. Since this quantity has to be conserved for the motion of the object at all times, assuming an ... 4 To a first approximation distance covered by the Moon is the same as the Earth's, but you can also estimate the correction to first order. Assume both orbits are circular and in the same plane since any deviations will affect only smaller order corrections. Represent the position in the orbital plane as a complex number$Z = R e^{2\pi i (t/Y)} + r ...

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