what is the distance between the planet and the star?

Awainaideannagi
2022-07-25
Answered

the distance between objects in space are so great that units other than miles or kilometers are often used. For example the astronomical unit (AU) is the average distance between earth and the sun or 92,900,000 miles. use this information to convert the distance of some planet in miles from its stars to astronomical unit. the planet is 893.1 million miles from the star.

what is the distance between the planet and the star?

what is the distance between the planet and the star?

You can still ask an expert for help

Lillianna Mendoza

Answered 2022-07-26
Author has **16** answers

92900000 mles = 1AU

893.1 *10^6 miles = ?Au

893.1*10^6/92900000 = 9.61 AU

893.1 *10^6 miles = ?Au

893.1*10^6/92900000 = 9.61 AU

asked 2020-12-30

A majorette in a parade is performing some acrobatic twirlingsof her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm.

With a skillful move, the majorette changes the rotation ofher baton so that now it is spinning about an axis passing throughits end at the same angular velocity 3.00 rad/s as before. What is the new angularmomentum of the rod?

asked 2022-05-18

When I moving object in some frame of reference, it has a positive kinetic energy. In the frame of reference where the object is stationary, it has a kinetic energy of 0.

Is the energy difference in some way associated to the frames of reference?

Is the energy difference in some way associated to the frames of reference?

asked 2022-05-20

Why is the de Broglie equation as well as the Schrodinger equation is correct for massive particle?

Starting from special relativity, here I see the de Broglie approximation is valid only if ${m}_{0}=0$

Derivation:

${E}^{2}={P}^{2}{C}^{2}+{m}_{0}^{2}{C}^{4}$. Here we put Plank-Einstein relation $E=h\nu =h\frac{C}{\lambda}$. Finally,

$\lambda =\frac{h}{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}\phantom{\rule{2cm}{0ex}}(1)$

If ${m}_{0}=0$ then $\lambda =\frac{h}{p}$ (de Broglie approximation).

Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if ${m}_{0}\ne 0$. But if we take special relativity very strictly then this approximation looks incorrect.

In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-

$\mathrm{\Psi}=A{e}^{i(\frac{2\pi}{\lambda}x-\omega t)}=A{e}^{i(\frac{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}{\hslash}x-\frac{E}{\hslash}t)}\phantom{\rule{2cm}{0ex}}$ (putting $\lambda $ from '1', $\frac{h}{2\pi}=\hslash $ and $E=\hslash \omega $).

Then, $\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=-\frac{{P}^{2}+{m}_{0}^{2}{C}^{2}}{{\hslash}^{2}}\mathrm{\Psi}=-\frac{{E}^{2}}{{C}^{2}{\hslash}^{2}}\mathrm{\Psi}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{C}^{2}{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}\phantom{\rule{4cm}{0ex}}(2)$

Again, $\frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}=-i\frac{E}{\hslash}\mathrm{\Psi}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\mathrm{\Psi}=-i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}$

Here we see operator $E=-i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{t}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}\phantom{\rule{4cm}{0ex}}(3)$

Combining (2) and (3) we find the differential equation:

$\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=\frac{1}{{C}^{2}}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}$

It is the Maxwell's equation, not the well known Schrodiner equation!

Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for ${m}_{0}\ne 0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

Starting from special relativity, here I see the de Broglie approximation is valid only if ${m}_{0}=0$

Derivation:

${E}^{2}={P}^{2}{C}^{2}+{m}_{0}^{2}{C}^{4}$. Here we put Plank-Einstein relation $E=h\nu =h\frac{C}{\lambda}$. Finally,

$\lambda =\frac{h}{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}\phantom{\rule{2cm}{0ex}}(1)$

If ${m}_{0}=0$ then $\lambda =\frac{h}{p}$ (de Broglie approximation).

Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if ${m}_{0}\ne 0$. But if we take special relativity very strictly then this approximation looks incorrect.

In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-

$\mathrm{\Psi}=A{e}^{i(\frac{2\pi}{\lambda}x-\omega t)}=A{e}^{i(\frac{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}{\hslash}x-\frac{E}{\hslash}t)}\phantom{\rule{2cm}{0ex}}$ (putting $\lambda $ from '1', $\frac{h}{2\pi}=\hslash $ and $E=\hslash \omega $).

Then, $\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=-\frac{{P}^{2}+{m}_{0}^{2}{C}^{2}}{{\hslash}^{2}}\mathrm{\Psi}=-\frac{{E}^{2}}{{C}^{2}{\hslash}^{2}}\mathrm{\Psi}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{C}^{2}{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}\phantom{\rule{4cm}{0ex}}(2)$

Again, $\frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}=-i\frac{E}{\hslash}\mathrm{\Psi}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\mathrm{\Psi}=-i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}$

Here we see operator $E=-i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{t}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}\phantom{\rule{4cm}{0ex}}(3)$

Combining (2) and (3) we find the differential equation:

$\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=\frac{1}{{C}^{2}}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}$

It is the Maxwell's equation, not the well known Schrodiner equation!

Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for ${m}_{0}\ne 0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

asked 2022-05-20

asked 2022-07-23

Does a photon in vacuum have a rest frame?

asked 2022-08-05

Why is it that when one considers some plane wave:

$\mathbf{E}(\mathbf{r},t)={\mathbf{E}}_{0}{e}^{i({k}_{0}z-{\omega}_{0}t)}$

$\mathbf{E}(\mathbf{r},t)={\mathbf{E}}_{0}{e}^{i({k}_{0}z-{\omega}_{0}t)}$

asked 2022-07-14

If you are traveling at $x$ speed the time will pass for you slower than to an observer that is relatively stopped. That's all just because a photon released at the $x$ speed can't travel faster than the $c$ limit.

What happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?

What happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?