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22

The main plot below shows the potential energy of a mass in the Earth-Moon system under the unrealistic assumption that the system is not rotating. i.e. This mirrors (at present) all but one of the 4 answers given, in assuming that this point is defined where the gravitational force on a mass due to the Earth and the Moon are equal and opposite (i.e. at the ...


16

Set the forces on the test particle from the Earth and Moon equal: $$F_E=F_M$$ $$G\frac{M_EM_{\text{ test particle}}}{R_E^2}=G\frac{M_MM_{\text{ test particle}}}{R_M^2}$$ The $G$s and $M_{\text{ test particle}}$s cancel, leaving you with $$\frac{M_E}{R_E^2}=\frac{M_M}{R_M^2}$$ but you know that $R_M$, the distance between the test particle and the Moon, is ...


11

Here is how I interpret what happened: You used Excel to compute the coefficients of the Gaussian that best describe the data: mean $\mu$, standard deviation $\sigma$, and magnitude $A$ for a curve $$Y=Ae^{-(x-\mu)^2/2\sigma^2}$$ Then you evaluated that function at a number of X values. Since the X values are not symmetrical about the calculated mean, you ...


9

At Lagrange point L1. Specifically for Earth-Moon L1, these calculations show 326054 km.


9

Earth is about 100x more massive than the moon, and since $F \propto M / r^2 $, the distance from Earth to the astronaut would have to be about $\sqrt{100}$ = 10x further than from the moon to the astronaut. Therefore, the astronaut falls "up" about 90% of the way to the moon. [The earlier answers go a lot more into detail (and are more technically ...


4

Hints: In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution $$\tag{1} ...


4

Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is ...


3

To calculate this by yourself, you need to know that gravity force exerted on an object (for exapmle You) is equal to $F=GMm/r^2$, where $G$ is gravity constant, $M$ is the mass of the big object ($M_m$ for moon, $M_e$ for earth), $m$ is the mass of small object. $r$ is the distance from the center of the mass. Now you need to know masses of earth and moon ...


3

Just use an equation derogating from the two forces which pull the objects (universal gravitation)to get the equilibrium point, something like (already simplifyed): M/d^2 = m/(384000000 - d)^2 Where M is the mass of earth, m the mass of moon and d the distance from earth. As d gets bigger than this value, you start falling into the moon I get a value of ...


3

The distance I got was 346 084km. Here are the maths I used: ($E_m$) Earth mass = $5.9736\times10^{24}$ kg ($M_m$) Moon mass = $7.3477\times10^{22}$ kg ($D_{em}$) average Earth-Moon distance = 384 467km ($G$) gravitational constant = $6.67384\times10^{-11}$ ($W$) my weight = 85kg ($D_{fe}$) distance from earth = ? The attraction force between two objects ...


3

This question is about Guassian wave-packet propagation and the corresponding Green's function in ordinary quantum mechanics. Assuming $\hbar=m=1$ for simplicity, consider the solution (with its initial condition) to the following Schrodinger equation: $$i\partial_t G=-\partial^2_x G \\ G(t=0,x)=\delta(x) $$ Now assume the Fourier ansatz for $\psi$: ...


3

Note that Fitzpatrick states towards the beginning, The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: ...


3

Ahh, Richard Fitzpatrick. Great guy. Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g. $$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...


2

It should be in joules. The term $\ln \frac{V_2}{V_1}$ is unitless, because the units on the top and bottom of the fraction cancel. Pressure, in pascals, is defined as Newtons per square meter. You've already put the volume into cubic meters, and the cancellation goes $$\text{meters}^3 \times \frac{\text{Newtons}}{\text{meters}^2}=\text{Newtons} \times ...


2

Every time, when you deal with differential equations, the first step is to put it into dimensionless form. There are more reasons for that. First, "small" and "large" has no meaning in dimensional forms, since you can always change the system of units. Second, nature knows no units. Now, when there is no exact solution (this is often the case), you can ...


2

If the block is resting on a frictionless surface, it won't fall over. It'll set in horizontal motion even with little force. To create torque, one edge of the bottom surface of the block needs to be fixed with high static friction and there must be a non-zero angle between the force and the line connecting force application point and fixed point (you can ...


2

You're right that in the context of radioactivity, antineutrinos are pretty much only released when a neutron turns into a proton, ${}_0^1n\to {}_1^1p+{}_{-1}^{\;0}e+\bar{\nu}$. They can also be consumed when a proton turns into a neutron and a positron, ${}_1^1p + \bar{\nu}\to {}_0^1n + {}_{1}^{0}e$. There are some other processes that involve ...


2

$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.


1

A gaussian fit is symmetrical by definition, because it is a gaussian. Your orange fit doesnt look like a gaussian, it is not even smooth. I do no think excel had a gaussian fit function (but I dont use excell so cannot tell for sure. You can use other software such as matlab, or likely free ones on the web. Or, just use that data to calculate the parameters ...


1

This proof from Griffiths book introduction to electrodynamics Consider the vector function $$\vec{a}=\frac{1}{r^2}\hat{r}$$ At every location $\vec{a}$ is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get ...


1

Start with ${b \over g }=m$ and rearrange this equation to get $b$ in terms of $g$ Then you can substitute your expression for $b$ into $g+b=3m$ and you should get the answer.... does this help?


1

Do some substitution: Re-arrange $$g+b=3m$$ to become $$b=3m-g$$ With substitution, $$\frac{b}{g}=10$$ becomes $$\frac{3m-g}{g}=10$$ Can you go from here?


1

Let the operator $q$ be defined by $f \mapsto \exp\left(i\,\frac{l}{\hbar}\,x\right)\,f$ then you should be able to prove to yourself the identity: $$p\,q\,f = q\,(p+l)\,f\tag{1}$$ for any vector (i.e. function) f. (This is just a variation on finding an integrating factor to transform $\mathrm{d}_x\,f + i\frac{l}{\hbar} f$ into the derivative of a product ...


1

Your answer seems to be good. I solved it in the same way. However there are two points in which i have no clarity. The first thing i do not have clear is the conservation of energy in the problem. We use the energy conservation, but how do we guarantee that the normal force is conservative? And, still, in the conservation equations we do not consider its ...


1

The answers in the comments are both correct - a couple of points to add. First think about the reverse process - how would you work out the energy required to break up the proton and electron in hydrogen to form a proton and electron. This is the same as the energy that needs to be released on formation of a hydrogen atom. I think you have figured this out ...


1

Within the specification I can glean from the question - here is what I would do. (i) Find the best fit Gaussian - which I am assuming is what you have done. (ii) Your best fit should return a chi-squared value You should compare the chi-squared value with critical values of the chi-squared distribution for the appropriate number of degrees of freedom of ...


1

You said that it's from a Brazilian formula sheet. I'm not sure if it's the correct translation, but Google calls work "trabalho" and kinetic energy "energia cinética". So it could be that $\tau$ refers to "trabalho"—work— and $E_c$ "energia cinética"—kinetic energy. So $\tau = \Delta E_c$ would be the work-kinetic energy theorem (work is the change in ...


1

First off, Strontium accumulates in the bones replacing Calcium, so you can't easily get rid of it. Second, the most radioactive natural food we eat is the banana, at about 130 becquerels per Kg. So that 1L of "hot" water really is quite nasty if you drink it and a significant portion of the Sr90 ends up in your bones.


1

The specific gravities would be the same if the levels of the two side were the same after liquid-II was added. When I try this, my logic seems to be flawed too. I don't get 1.12. The level on side II has not changed. The level on side I has risen 2 cm. So 2 cm of liquid-II were added. Consider the horizontal plane 2 cm below the top of side II. Below ...



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