| @@ -1,44 +1,31 @@ |
| | | ### Statement |
| | | |
| | − | $2.6.30.$ [Insert problem description here] |
| | + | $2.6.30.$ What speed should be given to a body of small mass in the center of an asteroid |
| | + | of mass $m$ and radius $R$, so that it goes infinitely far from the asteroid through |
| | + | a radial shaft? The asteroid can be considered homogeneous. |
| | | |
| | − | __Example Statement__: |
| | − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. |
| | + | #### Solution |
| | + | Important thing - in the center of an asteroid the body's gravitational potential energy will not be zero. Each layer of the asteroid will affect the body. We need to find the gravitational potential in the center for each layer: |
| | + | $$\phi_G=-\int\frac{\gamma dm}{r}$$ |
| | + | The mass of each layer we can find with common equation of density(due to the homogeneity of the asteroid) |
| | + | $$dm=\rho dV$$ |
| | + | Density of ball: |
| | + | $$\rho =\frac{3m}{4\pi R^3}$$ |
| | + | Derivative of the volume of a sphere with respect to its radius |
| | + | $$dV=4\pi r^2dr$$ |
| | + | let's substitute this into the potential equation |
| | + | $$\phi_G=-4\pi\gamma \rho\int rdr$$ |
| | + | integration is carried out in the range from zero to $R$ |
| | + | $$\phi_G=-4\pi\gamma \rho\frac{R^2}{2}$$ |
| | + | let's substitute the density |
| | + | $$\phi_G=-\frac{3\gamma m}{2R}$$ |
| | + | in order for the body to leave the gravitational fields of the asteroid, the amount of mechanical energy must be equal to zero |
| | + | $$M\phi+\frac{Mv^2}{2}=0$$ |
| | + | from here we can find the square of the velocity |
| | + | $$v^2=\frac{3\gamma m}{R}$$ |
| | + | hence the speed itself |
| | + | $$\boxed{v=\sqrt{\frac{3\gamma m}{R}}}$$ |
| | | |
| | − | |
| | − | ### Solution |
| | − | |
| | − | [Your solution should be placed here] |
| | − | |
| | − | __Example Solution__: |
| | − | The acceleration of the body defined by |
| | − | |
| | − | $$a(t) = bt$$ |
| | − | |
| | − | We know that acceleration is the time derivative of velocity: |
| | − | |
| | − | $$a(t) = \frac{d v(t)}{d t}$$ |
| | − | |
| | − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: |
| | − | |
| | − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ |
| | − | |
| | − | If the initial velocity is $v(0) = 0$, then the velocity becomes: |
| | − | |
| | − | $$v(t) = \frac{b t^2}{2}$$ |
| | − | |
| | − | Likewise, integrate $v(t)$ with respect to time: |
| | − | |
| | − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ |
| | − | |
| | − | From where the coordinate from time, considering the initial conditions: |
| | − | |
| | − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ |
| | − | |
| | | #### Answer |
| | | |
| | − | [Insert a concise answer or boxed result, like this:] |
| | − | |
| | − | |
| | − | __Example Answer__: |
| | − | $$ x(t)=\frac{bt^3}{6} $$ |
| | + | $$v=\sqrt{\frac{3\gamma m}{R}}$$ |
