Правка разделов «Solution», «Answer»
en/2.6.38.md
+5 −4
| ### Statement | |||
| $2.6.38.$ 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 | |||
| 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 | |||
| @@ -15,10 +15,11 @@Solution | |||
| $$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$ | |||
| 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 | ||
| − | $$\rho^2=v^2R^2+2\gamma MR$$ | ||
| + | let us express the square of the sighting parameter | ||
| + | $$\rho^2v^2=v^2R^2+2\gamma MR$$ | ||
| + | $$\rho^2=R^2+2\gamma MR/v^2$$ | ||
| and we will get an answer | |||
| − | $$\boxed{\rho=R\sqrt{1+2γM/(Rv2)}}$$ | ||
| + | $$\boxed{\rho=R\sqrt{1+2γM/(Rv^2)}}$$ | ||
| #### Answer | |||
| − | $$\rho=R\sqrt{1+2γM/(Rv2)}$$ | ||
| + | $$\rho=R\sqrt{1+2γM/(Rv^2)}$$ | ||
| ### Statement | ### Statement | ||
| $2.6.38.$ A space probe is moving towards a planet of radius $R$ and mass $M$ from a | $2.6.38.$ 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 | distance with a velocity $v$ relative to it. At what sighting parameter $ρ$ will the | ||
| probe fly closest to the planet without crashing? | probe fly closest to the planet without crashing? | ||
| #### Solution | #### Solution | ||
| First of all, we need to write angular momentum at the beginning and at the surface of the planet: | First of all, we need to write angular momentum at the beginning and at the surface of the planet: | ||
| $$mvsin\alpha* r=muR$$ | $$mvsin\alpha* r=muR$$ | ||
| We have triangle with the opposite side $\rho$ and hypotenuse $r$, so: | We have triangle with the opposite side $\rho$ and hypotenuse $r$, so: | ||
| $$rsin\alpha=\rho$$ | $$rsin\alpha=\rho$$ | ||
| Then we can express the speed at the surface | Then we can express the speed at the surface | ||
| $$u=\frac{v\rho}{R}$$ | $$u=\frac{v\rho}{R}$$ | ||
| We need to write law of conservation of energy | We need to write law of conservation of energy | ||
| @@ -15,10 +15,11 @@Solution | |||
| $$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$ | $$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$ | ||
| Let's substitute the speed we know at the surface | Let's substitute the speed we know at the surface | ||
| $$\frac{v^2R+2\gamma M}{2R}=\frac{v^2\rho^2}{2R^2}$$ | $$\frac{v^2R+2\gamma M}{2R}=\frac{v^2\rho^2}{2R^2}$$ | ||
| et us express the square of the sighting parameter | let us express the square of the sighting parameter | ||
| $$\rho^2=v^2R^2+2\gamma MR$$ | $$\rho^2v^2=v^2R^2+2\gamma MR$$ | ||
| $$\rho^2=R^2+2\gamma MR/v^2$$ | |||
| and we will get an answer | and we will get an answer | ||
| $$\boxed{\rho=R\sqrt{1+2γM/(Rv2)}}$$ | $$\boxed{\rho=R\sqrt{1+2γM/(Rv^2)}}$$ | ||
| #### Answer | #### Answer | ||
| $$\rho=R\sqrt{1+2γM/(Rv2)}$$ | $$\rho=R\sqrt{1+2γM/(Rv^2)}$$ | ||