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mariarad [96]
4 years ago
6

A lunar module (LM) lifts off from the lunar surface and flies a powered trajectory to its burnout point at 30 km altitude. The

velocity vector of the LM is parallel to the lunar surface at burnout. It then coasts halfway around the moon, where it must climb and rendezvous with the Apollo command module (CM) in a 250-km circular orbit. Note: μmoon = 4902.8 km3/s2, rmoon = 1740 km.a) Calculate vp, the burnout speed of the LM (km/s).b) Calculate vCM, the CM circular orbit speed (km/s).c) Calculate va, the LM orbit speed when it reaches the CM (km/s).d) Calculate the ∆v required to match speeds at the rendezvous with the CM (km/s).e) Calculate tcoast, the required coast time for the LM to reach the CM (seconds).f) In order to assure a rendezvous, it is desirable that the LM and CM arrive at the rendezvous point together. Where must the CM be in relation to the LM at burnout? Cite your answer as a time differential and an angle differential of the CM ahead of, or behind the LM burnout point. This sets the "launch window" for the LM takeoff.

Physics
1 answer:
xxTIMURxx [149]4 years ago
7 0

Answer:

Detailed solution is given below:

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Answer:

<h2> 27m/s</h2>

Explanation:

Given data

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The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ <em>F</em> = <em>N</em> - <em>W</em> = 0

• net horizontal force:

∑ <em>F</em> = <em>Fs</em> = <em>m a</em>

where

<em>N</em> = magnitude of normal force

<em>W</em> = car's weight

<em>Fs</em> = mag. of static friction

<em>m</em> = car's mass

<em>a</em> = <em>v</em> ²/<em>R</em> = mag. of the centripetal acceleration

<em>v</em> = car's speed

<em>R</em> = radius of curve

Now,

• compute the car's weight:

<em>W</em> = <em>m g</em> = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

<em>N</em> = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

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• solve for the (maximum) acceleration:

7350 <em>N</em> = (1500 kg) <em>a</em>   →   <em>a</em> = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = <em>v</em> ²/ (35 m)   →   <em>v</em> ≈ 13 m/s

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3 years ago
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