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shutvik [7]
3 years ago
11

A spacecraft returning from a lunar mission approaches earth on a hyperbolic trajectory. At its closest approach A it is at an a

ltitude of 5000 km, traveling at 10 km/s. At A retrorockets are red to lower the spacecraft into a 500 km altitude circular orbit, where it is to rendezvous with a space station. Find the location of the space station at retrore so that rendezvous will occur at B.
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

Explanation:

rA= 5000km+6378km=11378 km

rB= 500km+6378km=6878 km

Eccentricity, e=[11378-6878] / [11378 +6878] =0.24649

Evaluating the orbital equation at perigee yields the angular momentum

rB= h22/u * [1/[1+e]]----------------[1]

Pluging the values in equation[1]

6878 km= h22/398600* [1/[1+0.24649]]

we get h2= 58458 km2/s

Period of transfer ellipse, T_{2}=\frac{2\pi }{\mu ^{2}}( \frac{h_{2}}{\sqrt{1-e^{2}}})^{3}

Pugging the values we get , T_{2}=\frac{2\pi }{398600 ^{2}}( \frac{58458}{\sqrt{1-0.24649^{2}}})^{3}=\mathbf{8679.1s}

The period of circular orbit 3 is ,T_{3}=\frac{2\pi }{\mu }*r_{B}^{\frac{3}{2}}

T_{3}=\frac{2\pi }{398600 }*6878^{\frac{3}{2}}=\mathbf{5676.8s}

The time of flight from C to B on orbit 3 must be equal to  time of flight from A to B on orbit 2

\bigtriangleup t_{CB}=\frac{1}{2}T_{2}=\frac{1}{2}*8679.1=4339.5s

here orbit 3 is a circle, hence

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Answer:

curves downward, below the initial velocity vector.

Explanation:

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Regardless of the type, when reading the paragraph above, we can say that the trajectory of a projectile will always be curved down and below the initial velocity vector.

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D. Validity group

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A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg ma
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<em>A clamp-type measuring instrument operates on the principle of; </em>

A. induction

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