Respuesta :
Explanation:
Given that,
Distance = 402000 km
Speed = 2.23 m/s
Angle = 150
(a). We need to calculate the eccentricity of the trajectory
Using formula of eccentricity
[tex]\epsilon=\dfrac{v^2}{2}-\dfrac{\mu}{r}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{2.23^2}{2}-\dfrac{398600}{402000}[/tex]
[tex]\epsilon=1.4949\ km^2/s^2[/tex]
We need to calculate the angular momentum
Using formula of the angular momentum
[tex]h^2=-\dfrac{1}{2}\dfrac{\mu^2}{\epsilon}(1-e^2)[/tex]
[tex]h^2=-\dfrac{1}{2}\dfrac{(398600)^2}{1.4949}(1-e^2)[/tex]
[tex]h^2=-5.3141\times10^{10}(1-e^2)[/tex]...(I)
The orbit equation is
[tex]h^2=\mu r(1+e\cos\theta)[/tex]
[tex]h^2=398600\times402000(1-+\cos150)[/tex]
[tex]h^2=16.02372\times10^{10}(1-e0.8660)[/tex]
[tex]h^2=16.02372\times10^{10}-e13.877\times10^{10}[/tex]....(II)
Equating the value of h²
[tex]-5.3141\times10^{10}(1-e^2)=16.02372\times10^{10}-e13.877\times10^{10}[/tex]
[tex]-5.3141+5.3141e^2=13.877e+16.02372[/tex]
[tex]5.3141e^2+13.877e-21.33782=0[/tex]
[tex]e = 0, 1.086[/tex]
(b). We need to calculate the altitude at closest approach
Put the value of e in equation (I)
[tex]h^2=16.02372\times10^{10}-1.086\times13.877\times10^{10}[/tex]
[tex]h^2=9.53298\times10^{9}\ km^4/s^2[/tex]
Now, using the formula of the altitude at closest
[tex]r_{perigee}=\dfrac{h^2}{\mu}\dfrac{1}{1+e}[/tex]
[tex]r_{perigee}=\dfrac{9.53298\times10^{9}}{398600}\dfrac{1}{1+1.086}[/tex]
[tex]r_{perigee}=11465\ km[/tex]
So, The altitude is
[tex]z_{perigee}=r_{perigee}-r_{earth}[/tex]
[tex]z_{perigee}=11465-6378[/tex]
[tex]z_{perigee}=5087\ km[/tex]
(c). We need to calculate the speed at the closest approach.
Using formula of speed
[tex]v_{perigee}=\dfrac{h}{r_{perigee}}[/tex]
[tex]v_{perigee}=\dfrac{\sqrt{9.53298\times10^{9}}}{11465}[/tex]
[tex]v_{perigee}=8.516\ km/s[/tex]
Hence, This is the required solution.
