# Tag Info

1

The first equation is a very close approximation since m (satellite's mass) << M (Earth's mass) so m can be ignored. The second equation is the mathematically correct one.

1

Because of the centrifugal force. I had made some calculations back when I was looking for a good MSc thesis subject because I wanted to work on a railgun or a slingshot in orbit but it turned out it was impossible because of that, since it would need an astronomically long arm not to crush everything in the satellite. You are looking at reaching orbital ...

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Part of the IAU definition of a planet is that it "has 'cleared the neighbourhood' around its orbit". So, you can only have a single planet at a particular distance from its parent star as a matter of definition. Otherwise, it wouldn't be a planet.

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about the third law: I've read that Kepler was actually a mathematician. believing in heliocentric universe, Brahe's observations of planets some times deviated significantly from the place expected by the Copernican theory of circular orbits. strongly believing Brahe haven't made such great mistakes, Kepler thought, what if the orbits weren't circular???? ...

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Ok, I'm going to summarize the info from the comments. Assuming: (1) E = - G Mm / r^2 is the energy of an orbiting mass. from wikipedia G = 6.6.7 x10^-11 M = in this case, mass of sun = 1.9891 x 10^30 kg m = in this case, mass of earth, 5.97 x 10^22 r (earth) = radius of earth orbit = 149,598,261 km r (mars) = radius of mars orbit = 227,939,100 km ...

0

At a very basic level for the computation of a circular orbit it is just enough to equate the centripetal and the gravitational force: $$F_g=F_c$$ $$G \frac{mM}{r^2} = m \frac{v^2}{r}$$ where $G$ is the gravitational constant, $m$ is the mass of the satellite, $M$ is the mass of the Earth, $v$ is the satellites tangential velocity and $r$ is the altitude of ...

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First of all it is a bit strange to say that scientists place satellites into orbit. Since a rocket does all the work, which in turn is build by engineers. But you might say that the people who control the rocket/satellite can be called scientists. I am not an expert on the planning of trajectories of satellites. However I do suspect that the trajectories ...

1

Do you think anyone calculated the earth's speed to stay in orbit around the sun? As long as the speed is in the correct range the satellite will stay in orbit. For a satellite around the earth, the minimum speed is about 7 km/s. This is tangential speed, i.e. speed parallel to the earth's surface. Anything below 7 km/s, and the satellite will fall back. ...

2

The energy is still there in the form of gravitational potential. Think of leaving the earth as a process similar to riding a bicycle up a hill. (When you ride a bike up a hill, you're moving against Earth's gravitational field thereby gaining potential energy, just like what happens to the rock when it moves up away from the surface of the earth.) If the ...

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From a purely energy point of view, calculate the kinetic and potential (relative to the sun) energies of earth at Mars' orbit minus that in its current orbit. The kinetic energy is very straight forward, 1/2 m v2. Velocity (approximating to circular orbit) is only a function of the mass of the sun and the orbital radius. Or, you can simply look up the ...

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This answer is an complement to Chris White's answer. Fist of there is no explicit equations for the position of an object following a Kepler orbit as a function of time. However, when the initial conditions are known, the path the object will follow can be found, as well as the velocity, acceleration, ect. at any given position. This path can be described ...

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It seems you've done the hard part already, which is to evolve the object's position as a function of time. And moreover, the simulation seems stable over a number of orbits. (But eventually things start to go wrong; you may want to look at an answer I wrote to What is the correct way of integrating in astronomy simulations?) So my understanding is all you ...

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Orbital simulations can be handled by using the following relations: \begin{eqnarray} \mathbf F&=&m\mathbf a=m\frac{d^2\mathbf x}{dt^2}\tag{a} \\ \mathbf v&=&\frac{d\mathbf x}{dt}\tag{b}\\ \mathbf a&=&\frac{d\mathbf v}{dt}\tag{c} \end{eqnarray} The force acting on any two bodies, mass $M$ and $m$ is given by Newton's gravitational law ...

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When an object comes within the Roche limit, it breaks up because of tidal stresses - the part closest to the earth feels a stronger gravitational attraction than the furthest part. Hence, the closest part will fall a little faster than the trailing parts. As a result, "disintegration" does not mean that the body will fly apart like a bomb. Instead, it ...

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You may have noticed that if you start with the sun at rest, and put Jupiter into the system with an initial velocity to (say) the left, then over time the whole system moves left. (If you haven't noticed this is it worth setting the system up that way and letting it run long enough that you do notice it.) The trick is to recall that both bodies orbit their ...

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The moon orbits the earth with a near circular trajectory relative to the earth. So add earth's orbital velocity (around the sun) to the moon's orbital velocity (around the earth). This will put the moon into an orbit around the earth, but might make it a bit more eccentric (elliptical). To correct this you can use angular velocity around the sun with ...

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Here's how I picture this, w/out any calculations. Imagine the spaceship consisting of two massive objects, "front" and "rear" ones, connected with a long massless rod. They follow the same trajectory before and after interaction with the planet, with the "rear" one always a little behind the "front" one. The forces transmitted through the rod are always ...

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I assume there is only Earth and the ship in your system. For one thing the ship will not fall into orbit just like that. While falling, it gathers kinetic energy which will allow it to leave Earth again towards outer space, after its trajectory has been deflected by Earth gravity. The amount of deflection depends on its speed and how close it comes to ...

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