# Tag Info

## New answers tagged orbital-motion

1

I don't want to give away the answer directly. So I will provide some hints. A central force in polar coordinates has to be of the form: $$\vec{F} = m\vec{a} = m(\ddot r - r \dot \theta^2)\hat r$$ Now try to mess around with your $r(\theta) = a\theta^2$ I believe you need to specify $\dot \theta$ in order to solve the full equation of motion. So pick for ...

0

For a simple simulation you just need two equations that can be combined into one: Newton's law of universal gravitation: $$F=\frac{G\,m_1\,m_2}{r^2}$$ And Newton's second law: $$F=m\,a$$ So the final combined equation would be: $$a_1=\frac{G\,m_2}{r^2}$$ Or if your planets have positions $p_1$ and $p_2$ the vector equation would be: ...

3

My equation is $$\frac{x^2}{a^2} + \frac{x^2}{b^2} = 0$$ That's not the equation you want for a satellite. That equation describes an ellipse with its center at the origin. You want an ellipse with the origin at one of the foci: $$r = \frac{a(1-e^2)}{1+e\cos\theta}$$ where $r$ is the distance from the origin to a point on the ellipse. $a$ is the ...

2

1 is necessary for a geostationary orbit (which would be useful for an elevator location), but is not necessary for geosynchronous orbit (which your question asks). 2 is not advantageous. A (massive enough) satellite lower than geosynch altitude will induce tides that lower the altitude further, probably ending in a collision. Our moon is currently above ...

2

Tidal force acting on a natural satellite, like the moon around the earth, is the result of the deformability of the earth as the moon affects it and slowly the moon recedes from the earth. In general these tidal forces can be accelerating or decelerating : their orbital period is shorter than their planet's rotation. In other words, they revolve faster ...

1

Satellites in geosync are not "precisely positioned". Instead, they drift around and require station-keeping thrusters. If, by "tidal forces" you mean gravitational forces associated with the sun and the moon, then the answer is yes, and the effects are quite important.

1

The only obvious scenario involves placing a moon in the L2 point of a large enough planet. All the Lagrangian points except L4 and L5 are unstable. That implies that no such arrangement can persist without active station-keeping.

1

The gravitational potential energy is the energy stored in the gravitationnal field not in the masses them selves, the energy $E = mc^2$ is the mass-energy equivalence (at rest), think of it as energy has mass, most of the matter mass is due to the quark-gluon plasma energy and contributions from higgs mechanism,the more energetic particle the more massive, ...

7

Exerting a force and providing energy are quite different things. In particular, to provide energy to a body the force needs to perform work, that is, it needs to move the object in the direction that the force acts in. In the case of the Moon, the movement is circular and perpendicular to the gravitational force, so there is no inwards / outwards motion.* ...

2

Well, yes it is possible. Earth's rotational speed is decreasing. But the decrease rate is very small. The time period is increasing at a rate of nearly 2.2 seconds/ 100,000 years. So to increase the time period from 1 day to 365 days, it will require almost 1.4x10^12 years. This is calculated assuming that the rate of increase remains same which may not be ...

5

The escape velocity can be found in the usual way from the gravitational potential, but the potential no longer has the pure radial dependence you get with a spherically symmetric assumption. You have to compute the integral $$V = -G \iiint \!\!\mathrm{d}x \, \mathrm{d}y \,\mathrm{d}z \, \frac{\rho(x,y,z)}{R} \,,$$ where $\vec{R}$ is the displacement from ...

1

"As accurate as possible" is a fuzzy concept. Given that you ask this question, I expect that a few simplifying assumptions are justified. For an object in a circular orbit of constant radius $R$, orbiting a perfectly spherical earth of constant density, the kinetic and potential energy can be calculated. Their relationship is beautifully simple, as derived ...

1

Anything with mass will orbit around anything else with mass. Gravity is of infinite range - a proton on your nose knows about a proton in the Andromeda Galaxy. What's interesting is to see if the orbits produced mean anything sensible. This spreadsheet extract shows some orbital periods (T) for some common and hypopthetical pairs of objects. The point to ...

0

Although the force is radial, the direction of motion is not the direction of the force, rather it is the direction of the velocity at any time $t$. In order to find out the dependence $\mathbf{v}(t)$ one must solve the equations of motion $\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}})=m\mathbf{a}$. Doing so with the gravitational potential $V(r) = ... 0 This answer possibly isn't at the level that you would like, but I'm inclined to write it anyway because it's a good introductory answer. In the event that that this question does merge as duplicate, I'll probably just move my answer over there (as this is slightly different than the posted answers). First, watch this Minute Physics video as it provides a ... 0 Why can't you determine the angular momentum? If you have the velocity vector$\vec v$and the position vector$\vec r$, angular momentum can be calculated at any point in the orbit:$L=m \vec{r} \times \vec{v}$. In fact since you have the tangential velocity explicitly, it's even easier... 1 For a static black hole described by the Schwarzschild metric the escape velocity is: $$v_e = c\left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}} \tag{1}$$ and the orbital velocity is: $$v_o = c\sqrt{\frac{r_s}{2(r - r_s)}} \tag{2}$$ where$r_s$is the Schwarzschild radius: $$r_s = \frac{2GM}{c^2}$$ If we graph these we get: Note that the$x\$ ...

2

The difference between the observed precession rate and the calculated precession rate excluding GR is called the anomalous precession. The magnitude of the anomalous precession rate depends mostly on how close the orbit gets to the Sun because the curvature of spacetime decreases rapidly as we move away from the Sun. So we expect the anomalous precession ...

1

Since the question is not very precise I will consider only one case: Compass needle parallel to earth surface The orbits of the ISS look like this: It barely reaches the 50°latitude. Assuming that in 300-400 km above the sea level at this altitude the earth magnetic field lines still have a strong horizontal component, the compass needle should ...

2

A compass reacts to the strongest magnetic field. A compass would point to the magnetic north of the Earth unless you bring a magnet sufficiently close. As you get further away from Earth, its magnetic field weakens. However, the ISS is still sufficiently close to make the compass point towards the Earth's magnetic north, given that its plane is not ...

3

Yes. Under any normal operating mode today, the station rotates at the rate of once per orbit so that a particular part of the station always points at the earth. This is almost identical to the way the moon orbits the earth. But this is not something that has to happen. During assembly, there were times when the station did not rotate. So if there were ...

2

Once it goes into orbit, it will remain in the same attitude, what I mean is the cupola the astronauts look out of always looks "down," was far as I know. I don't think the ISS has any spin, so it should stay in the same position. The International Space Station orbits about 354 kilometers (220 miles) above the Earth and travels at approximately  ...

3

For missions beyond Earth orbit, it's very common to launch into a parking orbit around Earth before translunar or interplanetary injection. Therefore, the time of launch doesn't constrain the departure angle from Earth; you can get any departure angle you want by waiting no more than 90 minutes. This allows more flexibility in launch time even if the ...

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