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I think you are doing your math incorrectly if you get $10^{21}kg$. $$M = \frac{Rv^2}{G}$$ Let's try Jupiter from your reference. $$M = \frac{(778 \times 10^6km) (13.1\frac{km}{s})^2}{G}$$ $$M = \frac{(7.78 \times 10^{11}m) (1.31 \times 10^4 \frac{m}{s})^2}{6.6743 \times 10^{-11} \frac{m^3}{kgs^2}}$$ $$M = \frac{1.34 \times 10^{20} \frac{m^3}{s^2}}{6.6743 ... 8$$M_{gal}=Rv^2/G$$(G=6.67\times10^{-11} (N*[m/kg]^2) . Units of v and R are km/sec. and km., respectively) You gave G in MKS, then: R and v are m, m/s,  m= (\frac {1}{10^3}) km, that's why you got a wrong result:  10^3 * (10^3) ^2 = 10^9  that's the order of magnitude you are missing$$ 1.5*10^{11} *(3*10^4)^2/(6.6*10^{-11}) = ...

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An orbit in three dimensions is generally specified by giving how three quantities depend on time, or by giving how two coordinates depend on the third one.You have however omittied saying how $\phi$ depends on $\theta$; I will thus simply assume that $\phi$ is a constant, hence $v_\phi = 0$. In this case the total angular momentum $$\mathbf L = m\mathbf ... 0 Using differential geometry one can show that$$ \vec{p}(\theta) = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 a \sin \theta \cos \theta \\ 2 a \sin \theta^2 \\ 0 \end{pmatrix}  \vec{v}(\theta,\dot\theta) = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \frac{{\rm d}}{{\rm d}t} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = ...

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Using energy conservation you can find your answer from $$V = E-\frac{1}{2}mv^2$$ Express $V$ in terms of $r$ and you should get the form of the potential, assuming that it is a central force. Then you have to show that the assumption is correct. Is the problem such that you can assume $\varphi =0$?

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Using conservation of angular momentum you have (note that this relation implies a monotonic relation between angle and time) $$mr^2\dot{\varphi}=J\Rightarrow\mathrm{d}t=\frac{mr^2}{J}\mathrm{d}\varphi.$$ The energy is conserved and given by, $$E=\frac{1}{2}m\dot r^2+\frac{J^2}{2mr^2}+V(r).$$ Changing variables from time to angle ($\dot r=r_\varphi ... 0 You have four big problems and two small problems. The big problems are that you are initializing the Earth's and Moon's initial position and velocity incorrectly. The initial distance between the Earth and Moon is off by a factor of 1.0123, as is the initial relative velocity. The small problems are (1) an incorrect value for the Earth-Moon semi-major axis ... 1 The mean motion$n$of a satellite is its angular velocity, averaged over one period. In other words, if the satellite rotates around the Earth with period$P$, its mean motion$n$is $$n = \frac{2\pi}{P}$$ If the Earth were a perfectly spherical symmetric object, and there were no other perturbing agents in the Universe (in other words, if Earth + ... 0 The harmonic mean involves averaging the reciprocals, so it is dominated by the smaller of the two values. The reduced mass has the same requirement: The orbit of a small body around a large body is almost the same as the orbit of a small body about a fixed point. This also "explains" the factor of 2: in the limiting case m1>>m2, the reduced mass = m2, not ... 0 The FRW spacetime can be written as${\bf R} \times {\bf \Sigma}$where${\bf R}$where${\bf R}$represents the time direction and${\bf \Sigma}$is a homogeneous and isotropic three-manifold representing the spatial dimensions. This means that for every time there is a corresponding hypersurface for space. To help with the visualisation of hypersurfaces ... 5 An interesting way to answer Part 2: Using the angular velocity version of the centripetal force equation:$$F_c=m\omega^2r=\frac{GMm}{r^2}$$If we assume that$r$is both the orbital radius and the radius of the sphere being orbited and that the density of that sphere is$\rho$, then:$$M=\frac43\pi \rho r^3$$ then the equation ... 10 tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ... 2 Some key reading if you want to understand this stuff is chapters two to five of the IERS Technical Note 36, the IERS Conventions (2010). It's not just the J2000/FK5 frame (aka the EME2000 frame) that is associated with some epoch date. Every Earth-centered inertial frame has some epoch date. There are two fundamental reasons why this must be the case: ... 1 The other two answers provided so far are both very good, but are perhaps a bit too advanced for what I suspect is an introductory calculus-based physics question. So, how to understand this from a an introductory calculus-based physics level as opposed to an upper-undergraduate / graduate physics level? The most advanced concept I'll use is the vector ... 0 Over long periods gravitational interactions between planets can have a very large effect on their orbits - especially if the planet you are interacting with is Jupiter. Orbits can become "adjusted" into periods which are in resonance with Jupiter's period. It is also possible for a planet to be moved to a completely different part of the Solar System by ... 1 Write the cross product as a matrix equation: instead of${\rm d}_t \vec{r} = \vec{\omega}\times \vec{r}$we can write: $${\rm d}_t \vec{r} = \left(\begin{array}{ccc}0&-\omega_z&\omega_y\\\omega_z&0&-\omega_x\\-\omega_y&\omega_x&0\end{array}\right)\,\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{1}$$ (check that this is the ... 0 Assuming that$\bf{c}$is the angular velocity vector the motion of any particle that is part of rigid body in general is $${\bf v } = h {\bf c} + {\bf c} \times {\bf r}$$ where$h$is the motion pitch and${\bf r}$is the location of the particle. See this post on details of how the screw motion of a rigid body defined. The first part is not position ... 3 You've asked a very entertaining question, and the answer is not simple. Let's ignore collisions for the moment. The "purest" effect, that is, the one which involves no change on the part of the planet or its sun, is the effect of tidal bulges in the sun. Just as the earth, for instance, is not a perfect sphere due to tidal forces, so the sun is not a ... 1 Just remember that a Black Hole doesn't have infinite gravity - it just has however much mass created it in the first place. Yes, anything that gets within the event horizon is trapped forever, but that event horizon will actually be smaller than the size of the equivalent amount of mass composed of ordinary matter. This is also why "microscopic black holes" ... 2 I suspect that you made a mistake at step 5, since the only thing that is reversed is the radius with respect to the velocity, but the trajectory does look like an ellipse. You probably switched a plus sign with a minus sign in the denominator of the following equation, which relates the true anomaly to the radius,$$r = \frac{a(1-e^2)}{1 + e \cos ... 1 This is a very late answer, but no answer has been selected yet. Why do only big rocks (planets) have satellites, and not small ones? Almost 100 asteroids have satellites, and some of them are rather small. For example, asteroid 54509 YORP has a mass of only 1×1010 to 4×1010 kilograms. That sounds like a lot, but that's a body with a mean radius of ... 1 If you did extend your arm out, you would have indeed changed your angular momentum, and for angular momentum to be conserved your orbit would change slightly in the opposite direction. In addition (I am going by your theory of "partial Spaghettification,"), if your orbit was not thrown off by the gravitational pull of the black hole(which it would be, I ... 0 It actually would be possible to achieve an almost stable orbit IF you could make a trebuchet powerful enough to fire a projectile to the moon. You could get the projectiles to orbit several times at least. And as people pointed out, just making one to fire into space is practically impossible, but assuming you got past that step, you have another problem. ... 1 Both masses$m$are at a distance$R$of the center of mass. The distance between the two masses is then$2R$. To calculate the potential energy, consider the first mass fixed. The potential energy of the second mass is$\gamma \frac{m^2}{2R}$, with$\gamma$the gravitational constant. This is also the potential energy of the entire system. The kinetic ... 2 It is really rotating, with a period of one month, according to a Foucault pendulum. That's the prediction of Newtonian gravity, and GR's prediction can't diverge much from that when the gravitational field is so weak. (Also consider that a Foucault pendulum on Earth empirically measures the length of the sidereal day, not the solar day.) 1 The moon rotates once every 27 days (roughly). In order to keep one side facing the Earth, it needs to rotate. If it didn't rotate, it would look like it was rotating backwards as it orbited the Earth. 0 Because this is a two particle system, you can do this exactly. For a given mass$m$and distance between objects$2r$, you can compute the required angular velocity$\omega$(to remain in stable orbit) and thus the kinetic energy ($\frac12m(r\omega)^2\$). For that same configuration, you can compute the gravitational energy by seeing how much it takes to ...

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The reason why we don't see dust particles orbiting comparably tiny rocks within the solar system is just that the rest of the solar system interferes due to how slow such an orbit has to be: the presence of other massive bodies perturbs the orbits of dust particles. the solar radiation deorbits particles smaller than about 1mm. the solar radiation ...

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Short, no-math answer: Objects in space can have any orbit they want to have. How long they stay in that orbit depends on what else is out there. Over the past 5+ billion years the various interactions have sorted themselves out and we see what's left. There's no reason a planet could not be in a retrograde orbit, a perpendicular orbit, or a figure-eight ...

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