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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 ...


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}) = ...


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 ...


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 ...


2

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 ...


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: ...


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

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" ...


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 ...


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

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 + ...


1

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 ...


1

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 ...


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 ...


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 ...



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