The Space Shuttle does not directly inject into its prescribed mission orbit. Rather, themain engine shutdown occurs at the apog
ee of an elliptical orbit with 100-km altitudeand the external tank is discarded. Perigee of this orbit is at 30-km altitude. After thetank is released, the Shuttle uses its orbital maneuvering system (OMS) to maneuverinto an elliptical orbit with perigee at 100-km altitude and apogee at 250-km altitude.This is called the OMS-1 burn. At the new apogee, the Shuttle performs the OMS-2burn to circularize into the 250-km altitude mission orbit. Calculate:
(a) the burnout speed at apogee in the external tank disposal orbit;(b) the ΔV required for the OMS-1 maneuver;(c) the ΔVrequired for the OMS-2 maneuver;(d) if launching from the Kennedy Space Center, without a plane change maneuver,what is the range of inclinations possible for the Shuttle mission orbit? Why?
(a) the burnout speed at apogee in the external tank disposal orbit;(b) the ΔV required for the OMS-1 maneuver;(c) the ΔVrequired for the OMS-2 maneuver;(d) if launching from the Kennedy Space Center, without a plane change maneuver,what is the range of inclinations possible for the Shuttle mission orbit? Why?
1 answer:
Answer:
(a) burnout speed at apogee in the external tank disposal orbit=7.82 km/sec
(b) ΔV required for the OMS-1 =0.066 km/sec
(c) ΔV required for the OMS-2 =0.045 km/sec
(d) This part of the question is explained in detailed way in the attached file.
Explanation:
Detailed explanation of all the parts of the question are given in attached files.
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Answer: The frequency heard will be f = 275.675Hz
Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.
The formula to calculate the Doppler effect is as follows
f = () · f₀
f is the observed frequency;
c is the speed of sound;
Vs is velocity of the source;
f₀ is the emitted frequency of source;
Substituting and calculating,
f = · 300
f = 275.675 Hz
Thus, the frequency heard by the police officer is 275.675Hz.
Answer:
Because the gravitational attraction of the Sun hold them in motion around it
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r is the average distance between the Sun and the planet
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