Правка разделов «Statement», «Solution», «Answer»
en/2.6.38.md
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| @@ -1,44 +1,24 @@ | |||
| ### Statement | |||
| − | $2.6.38.$ | ||
| + | $2.6.38.$ A space probe is moving towards a planet of radius $R$ and mass $M$ from a | ||
| + | distance with a velocity $v$ relative to it. At what sighting parameter $ρ$ will the | ||
| + | probe fly closest to the planet without crashing? | ||
| − | |||
| − | $1.1.1.$ 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 | ||
| + | First of all, we need to write angular momentum at the beginning and at the surface of the planet: | ||
| + | $$mvsin\alpha* r=muR$$ | ||
| + | We have triangle with the opposite side $\rho$ and hypotenuse $r$, so: | ||
| + | $$rsin\alpha=\rho$$ | ||
| + | Then we can express the speed at the surface | ||
| + | $$u=\frac{v\rho}{R}$$ | ||
| + | We need to write law of conservation of energy | ||
| + | $$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$ | ||
| + | Let's substitute the speed we know at the surface | ||
| + | $$\frac{v^2R+2\gamma M}{2R}=\frac{v^2\rho^2}{2R^2}$$ | ||
| + | et us express the square of the sighting parameter | ||
| + | $$\rho^2=v^2R^2+2\gamma MR$$ | ||
| + | and we will get an answer | ||
| + | $$\boxed{\rho=R\sqrt{1+2γM/(Rv2)}}$$ | ||
| − | |||
| − | ### 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__: | ||
| − | $$ | ||
| + | $$\rho=R\sqrt{1+2γM/(Rv2)}$$ | ||
| @@ -1,44 +1,24 @@ | |||
| ### Statement | ### Statement | ||
| $2.6.38.$ |
$2.6.38.$ A space probe is moving towards a planet of radius $R$ and mass $M$ from a | ||
| distance with a velocity $v$ relative to it. At what sighting parameter $ρ$ will the | |||
| probe fly closest to the planet without crashing? | |||
| #### Solution | |||
| $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | First of all, we need to write angular momentum at the beginning and at the surface of the planet: | ||
| $$mvsin\alpha* r=muR$$ | |||
| We have triangle with the opposite side $\rho$ and hypotenuse $r$, so: | |||
| $$rsin\alpha=\rho$$ | |||
| Then we can express the speed at the surface | |||
| $$u=\frac{v\rho}{R}$$ | |||
| We need to write law of conservation of energy | |||
| $$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$ | |||
| Let's substitute the speed we know at the surface | |||
| $$\frac{v^2R+2\gamma M}{2R}=\frac{v^2\rho^2}{2R^2}$$ | |||
| et us express the square of the sighting parameter | |||
| $$\rho^2=v^2R^2+2\gamma MR$$ | |||
| and we will get an answer | |||
| $$\boxed{\rho=R\sqrt{1+2γM/(Rv2)}}$$ | |||
| ### Solution | |||
| [Your solution should be placed here] | |||
| __Example Solution__: | |||
| The acceleration of the body defined by | |||
| $$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}}$$ | |||
| #### Answer | #### Answer | ||
| $$\rho=R\sqrt{1+2γM/(Rv2)}$$ | |||
| [Insert a concise answer or boxed result, like this:] | |||
| __Example Answer__: | |||
| $$ |
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