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Vikki [24]
3 years ago
14

You are the pilot of a spaceship (9.1835 × 104 kg) and you enter into elliptical orbit around an uncharted planet, Rosseforp. At

periapsis (closest approach to the planet), your ship is 4.5 × 107 m above the surface of the planet; at apoapsis (most distant) it is 2.257×108 m above the surface. The mass of Rosseforp is found by your ship’s equipment to be 1.96×1025 kg, and the period of your ship’s orbit is 83.37 hours
Find the following:i. the semimajor axis of your ship’s orbit,ii. the radius of planet Rosseforp,iii. the eccentricity of your ship’s orbit,iv. the speed of your ship at perihelion and aphelion,

Physics
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

Check the explanation

Explanation:

The orbital period of a satellite that is given as (T) and the mean distance from the central body (R) are connected by the following equation: representing T as the satellite period, R will be represented as the average radius of orbit for the satellite (which is the distance from center of central planet)

Kindly check the attached image below to see the step by step explanation to the above question.

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A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
The Doppler Effect means that all observers of a moving wave source detect the same wave frequency.
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3 years ago
A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H
svetlana [45]

Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

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The horizontal distance or range of a projectile is given as;

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t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

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3 years ago
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Answer:

Explanation:

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