Mass loss rate

The rate of mass loss is a quantity that indicates the loss of mass of a body per time ( mass flow ). It is often used in astrophysics in connection with stars and star-like objects.

description

The rate of mass loss by which the matter decreases within the volume is obtained by integrating the matter flow which flows over the limiting surface : ${\displaystyle {\dot {M}}}$ ${\displaystyle O}$ ${\displaystyle {\vec {j}}}$ ${\displaystyle \partial O}$

${\displaystyle {\frac {\mathrm {d} M}{\mathrm {d} t}}\equiv {\dot {M}}=\int _{\partial O}{\vec {j}}\cdot \mathrm {d} ^{2}r<0}$

With

• the crowd ${\displaystyle M}$
• currently ${\displaystyle t}$
• the location coordinate . ${\displaystyle r}$

The reason for the integration is that it is, however , for the spatial component to be fixed. ${\displaystyle {\vec {j}}\sim {\vec {r}},t}$ ${\displaystyle {\dot {M}}\sim t}$ ${\displaystyle r}$

The mass loss rate is negative in relation to the reference volume. ${\displaystyle O}$

unit

For ease of use, the mass loss rate is usually given in the following units:

• Kilograms / second
• Tons / second or tons / hour (in publications)
• Solar masses / year (in stellar astrophysics).

generalization

The loss rate is generally a quantity which reflects the change in a physical quantity in a certain volume . Depending on how one chooses, the equation can also be used for other purposes: ${\displaystyle {\vec {j}}}$ ${\displaystyle O}$ ${\displaystyle {\vec {j}}}$

• one chooses z. B. as an electrical current density , so the charge loss rate can be determined analogously . ${\displaystyle {\vec {j}}}$
• one chooses z. B. as particle flux density , so the particle loss rate can be determined analogously (see. The following example). ${\displaystyle {\vec {j}}}$

example

At a detector we measure a particle flow of

${\displaystyle {\vec {j}}=j\cdot {\vec {n}},}$

With

• ${\displaystyle j}$ in unity ${\displaystyle \mathrm {\frac {Teilchen}{m^{2}\cdot s}} }$
• the normal vector which is perpendicular to our detector surface and at the same time points in the direction of a source (e.g. our sun). ${\displaystyle {\vec {n}},}$

If the design of the detector ensures that it only receives solar particles , then the value of the current of these ponds is exactly the value of the mean distance of the sun from the earth (the current density is inversely proportional to the square of the distance from the source :) . ${\displaystyle j}$ ${\displaystyle r=1{,}50\cdot 10^{11}\,\mathrm {m} }$ ${\displaystyle j\sim {\frac {1}{r^{2}}}}$

The surface of a sphere with radius r around the center of the sun is:

${\displaystyle \partial O=4\pi r^{2}=2{,}83\cdot 10^{23}\,\mathrm {m^{2}} }$

The sun's particle loss rate can then be calculated by multiplying the current density of the emitted particles by the surface area of ​​the sphere around the sun:

${\displaystyle \int _{\partial O}{\vec {j}}\cdot \mathrm {d} ^{2}r=\int _{0}^{2\pi }\int _{0}^{\pi }j\cdot r^{2}\cdot \mathrm {d} \theta \cdot \mathrm {d} \phi =2{,}83\cdot 10^{23}\,\mathrm {m^{2}} \cdot j\,\mathrm {\frac {Teilchen}{m^{2}\cdot s}} }$