52

Yes, but undetectably. The Earth-Sun system radiates a continuous power average of about 200 watts as gravitational radiation. As Wikipedia explains, “At this rate, it would take the Earth approximately $1\times 10^{13}$ times more than the current age of the Universe to spiral onto the Sun.” The Hulse-Taylor binary (two neutron stars, one a pulsar) was the ...


20

Yes, two bodies orbiting each other like this will indeed emit gravitational waves, regardless of whether or not they're compact objects like neutron stars or black holes. Obviously, most exoplanets will not emit strongly; a planet-star system generally involves large separations and non-relativistic speeds. Therefore, as G. Smith noted, while all such ...


12

G.Smith and HDE 226868 gave good answers. I would add that, in the Solar system case, the gravitational waves are clearly not the dominant factor in changing the (Keplerian parameters of) orbits. Momentum exchange between planets, solar radiation pressure, solar wind effects, tidal effects - every one of these (and probably more that I cannot recall right ...


7

As noted in @G.Smith's answer, Wikipedia gives a figure of $\sim 200 \, \mathrm{W}$ for Earth/Sol orbital radiation. Wikipedia didn't clearly cite the source, but this PDF is cited not long thereafter and may be it. That PDF claims that the radiated energy for a non-relativistic binary system is about $$ \frac{\mathrm{d}E}{\mathrm{d}t} ~=~ - \frac{32 \, G^...


6

NOTE: This answer is for the OP’s original question, which was about two masses. The question was completely changed after this answer was posted. To get an ellipse with semi-major axis $a$, the initial relative speed should be $$v=\sqrt{G(m_A+m_B)\left(\frac2r-\frac1a\right)}.$$ This is known as the vis-viva equation and is an expression of energy ...


4

Here is an algorithm you can implement. Let the small orbiting body have mass $m$. You know its initial position and velocity vectors, $\mathbf{r}_0$ and $\mathbf{v}_0$, in an arbitrary Cartesian coordinate system. I’ll assume that you want to treat the “significantly heavier” body with mass $M$ as being stationary and consider it to be the origin of your ...


3

One convenient and easy-to-understand choice of units would be to use the solar mass ($\text{M}_\odot$) as the unit of mass, the astronomical unit (au) as the unit of length, and the year (a) as the unit of time. You can convert the SI value of the gravitational constant into these units and discover that it has the value $G=39.48\text{ au}^3\text{M}_\odot^{...


3

Your calculation of $E_{tot}$ is correct. However, acc. Wikipedia: The Little Boy atomic bomb dropped on Hiroshima on August 6, 1945, exploded with an energy of about 15 kilotons of TNT (63 TJ) So that was $63 \times 10^{12}\text{ J}$. We get: $$\frac{1.49 \times 10^{13}}{63 \times 10^{12}}=0.24$$ which appears a more reasonable result.


3

Planes intersect along a line. Normally we see a full Moon shining in Her fullness for exactly the reason you cite, but just sometimes the Moon passes through an intersection point exactly when occluded by the Earth and so we get a lunar eclipse. Swap the rôles of Moon and Earth and there's your answer.


2

Nice question. By re-entering atmosphere spaceship transfers it's kinetic energy to impacting air molecules, thus molecules starts to move at a spaceship speed too. Knowing average molecules speed (which would be an aircraft speed) we can calculate shock-wave zone gas temperature : $$ T={m{\overline {v^{2}}} \over 3k_{B}} $$ Typical re-entry speed of Space ...


2

The deduction of the expression for $\vec{r} \cdot \dot{\vec{r}}$, where $\vec{r}$ is the displacement of the center of mass of the orbiting entity from the center of mass of the entity which it orbits and $\vec{v} := \dot{\vec{r}}$ is the velocity of the orbiting entity, is critical to the vector analysis which usually accompanies an orbital mechanics study....


2

Edit: As requested, here is a calculation for the ellipsis. An ellipsis can be parametrically described by $\vec r(t) = (a\cos \varphi(t),\, b\sin\varphi(t))^\top$ where $0\leq\varphi(t) < 2\pi$. Then, $$\vec v(t) = \dot{\vec{r}}(t) = (-a\dot{\varphi}(t)\sin\varphi(t),\, b\dot{\varphi}(t)\cos\varphi(t))^\top.$$ Now take the dot product $$ \vec{r}(t)\cdot\...


2

if you have this kind of differential equations: $$\vec{\ddot{r}}=-\vec{F}(\vec{r})\tag 1$$ you can get the Hamiltonian. multiply equation (1) from the left with $\vec{\dot{r}}$ $$\vec{\dot{r}}\cdot \vec{\ddot{r}}=-\vec{\dot{r}}\cdot\vec{F}(\vec{r})$$ thus: $$\frac{1}{2}\frac{d}{dt}(\vec{\dot{r}}\cdot \vec{\dot{r}})= -\vec{\dot{r}}\cdot\vec{F}(\vec{r})$$...


2

Your estimate of the energy to put the ISS into orbit seems quite reasonable, about 15 TJ. When it is on the Earth's surface it already has some rotational energy (it is moving at about 465 m/s) which will reduce the needed increase in kinetic energy a little, but the effect is not large. Your estimate of the energy yield of the Hiroshima bomb is a little ...


2

This is a typical exercise in calculus so I would recommend brushing up. Given your equation: $$v^2 = GM\left( \frac{2}{r} - \frac{1}{a}\right)$$ you want to ask how the RHS changes when you change the LHS. I.e. you apply the differential operator $d$. $$\frac{d(v^2)}{dv} = 2v \iff d(v^2) = 2vdv$$ $$\frac{d}{da} GM\left(\frac{2}{r} - \frac{1}{a}\right) = \...


2

Take the Sun(mass M) and the planet(mass m) to be point masses. Let the Sun be fixed at the origin and the planet be moving in the x-y plane, initial velocity of the planet be $v_o\hat{j}$ and initial position of the planet be $r_o\hat{i}$. At any given instant, let the planet's position, velocity and acceleration be $\vec{r}, \vec{v}, \vec{a}$ respectively. ...


1

$E$ is monotonic like $t$. It isn’t limited to be between $0$ and $2\pi$.


1

When you calculate the orbital eccentricity in terms of energy and angular momentum, the same formula applies for circular orbits, elliptical orbits, parabolic trajectories, and hyperbolic trajectories. That formula is $$e=\sqrt{1+\frac{2\epsilon h^2}{\mu^2}}.\tag1$$ Here $\epsilon$ is the specific orbital energy (the energy -- kinetic plus potential -- ...


1

If you had an interferometer that was accurate enough, you would be constantly in an ocean of gravitational waves. The frequencies of the planet's waves would be very low frequencies, about 1 period per year! Jupiter going to aphelion would vary in amplitude every 12 years. For the moment, 20Hz is the record for low frequency gravitational wave detection.


1

The vis-viva equation holds for parabolas and hyperbolas, so yes, you can use it to obtain an initial velocity if you also know $r$, the distance to that initial point. However, you would not be able to directly use the equation for the semimajor axis, $a$, that you provided. Plugging in $e = 1$ or $e > 1$ to that equation does not really give a result ...


1

The Lorenz system of equations come from a model of convection of a two-dimensional flow of fluid of uniform depth, where they represent the growth of particular wave-modes. They might apply to some convective systems, but they are very abstract. They have nothing to do with planetary orbits. That said, strange attractors as demonstrated by this system ...


1

They are not in the same plane, but the planes intersect, and the relative orientation goes through 360 degrees every 365.2524 days. The objection would be valid if the lunar plane were parallel to the earth's plane, but offset by more than one Earth radius, but that is not how it works. By the way, the alignment is call "syzygy".


1

Orbital mechanics are weird. So, (hypothetically) we have the Earth travelling in a perfect circular orbit around the Sun. The radius of that orbit, the velocity of the earth, and the mass of the Sun are all fixed, and satisfy the relationship cited in the question. Now the Sun loses a small amount of mass. The Earth is still traveling at the same ...


1

Welcome to Physics.StackExchange! Let $M(t)$ be the solar mass and $m$ the Earth mass. Assuming circular orbit with radius $r(t)$, The gravitational force between them is $$ F = \frac{\gamma mM(t)}{r^2} = ma = \frac{mv^2}{r} \Rightarrow \sqrt{\frac{\gamma M(t)}{r}} = \frac{dr}{dt}, $$ where the orbital velocity is $ v = dr/dt$. Further assume the mass of ...


1

Consider Kepler's Third Law: $$T^2=\Big(\frac{4 \pi^2}{GM}\Big)r^3$$ where $T$ is the orbital period and $r$ the orbit radius (for a circular orbit). So: $$T^2 \propto r^3$$ where $\frac{4 \pi^2}{GM}$ is the proportionality constant. Since as it depends on $M$, the ratio $\frac{T^2}{r^3}$ will vary accordingly.


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