Line of sight on surface of Moon

LINE OF SIGHT ON SURFACE OF MOON

INTRODUCTION

The Moon is a natural satellite of our planet earth which has a mean radius of 1737 km. Although the equatorial radius is not equal to its polar radius, we can approximate the Moon as a sphere. Any object on the surface of sphere has a finite view due to the curvature of the sphere. Thus any one can view only up to a finite distance before the horizon. The horizon is itself defined on the height of the object, the greater the height the more the view. In this article, we intend to determine the line of sight for an average human being on the surface of Moon.

ASSUMPTIONS

1.      The surface of Moon is smooth

2.      Moon is a homogeneous sphere

3.      Space time around Moon is not curved but flat

4.      The observer is at ground level

CALCULATION

 Fig .1

From figure 1,

R – Radius of Moon [m]                                                                                                                           

R = 1737 Km = 1737000 m (Eq. 1)

h – Height of the observer [m]

h = 5 feet

   = 1.5 m (Eq. 2)

                                                                                                                               {⸪ 1 feet = 0.3 m}

d – Observable distance by observer [m]                                                                                         

We can apply Pythagorean Theorem, 

d² = (R+h)² – R² (Eq. 3)

d² = 2Rh + h²

d = √ (2Rh+h²) (Eq. 4)

Now substitute equations (1), (2) in equation (4)

d = √ (2*1737000*1.5+1.5²)

d = 2282.7619 m

d = 2.2827 Km = [1.4175 miles]

This is the distance that can be viewed by an observer on the surface of Moon provided there is no dust storm.

CONCLUSION

We thus determined the line of sight or field of view for an observer on the surface of Moon.

Line of sight on surface of Earth

LINE OF SIGHT ON SURFACE OF EARTH

INTRODUCTION

Earth is the third planet in our solar system which has a mean radius of 6371 km. Although the equatorial radius is not equal to its polar radius, we can approximate the earth as a sphere. Any object on the surface of sphere has a finite view due to the curvature of the sphere. Thus any one can view only up to a finite distance before the horizon. The horizon is itself defined on the height of the object, the greater the height the more the view. In this article, we intend to determine the line of sight for an average human being on the surface of Earth.

ASSUMPTIONS

1.      The surface of Earth is smooth

2.      Earth is a homogeneous sphere

3.      The sky is clear and vision is not obscured

4.      Light does not undergo diffraction and refraction

5.      Space time around Earth is not curved but flat

6.      The observer is at sea level

CALCULATION

Figure.1 Kansas flat plain with maximum line of sight

Figure.2 Line of sight on Earth

From figure 2,

R – Radius of Earth [m]                                                                                                                            

R = 6371 Km = 6371000 m (Eq. 1)

h – Height of the observer [m]

h = 5 feet

   = 1.5 m (Eq. 2)

                                                                                                                               {⸪ 1 feet = 0.3 m}

d – Observable distance by observer [m]                                                                                         

We can apply Pythagorean Theorem,

d² = (R+h)² – R² (Eq. 3)

d² = 2Rh + h²

d = √ (2Rh+h²) (Eq. 4)

Now substitute equations (1), (2) in equation (4)

d = √ (2*6371000*1.5+1.5²)

d = 4371.8419 m

d = 4.3718 Km [2.7148 miles]

This is the distance that can be viewed by an observer on the surface of Earth provided the weather is clear.

CONCLUSION

We thus determined the line of sight or field of view for an observer on the surface of Earth.

Doppler shift underwater Pt8

DOPPLER SHIFT UNDERWATER PT8

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when one dolphin is chasing the other in a linear path. Consider two dolphins, dolphin2 [observer] moving at 12 mph chasing dolphin1 [source] which is moving at 8.3 mph. We’ll determine the apparent frequency as registered by dolphin2 when dolphin1 emits sound waves.

ASSUMPTIONS

1. Water has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of pressure is negligible

3. The amplitude of sound is unity

4. The water molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1) 

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in water at standard temperature and pressure conditions {V = 1531 m/s}

V_o – Velocity of dolphin2 [observer] (m/s)

V_s – Velocity of dolphin1 [source] (m/s)

The Doppler shift equation in this case is,

f’ = f₀*{[V + V_o]/[V + V_s]} (Eq. 2)

The ‘+’ sign in the numerator of equation (2) indicates that the observer is moving toward the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

  

The velocity of dolphin1 V_s = 8.3 mph

                                              = 3.68 m/s (Eq. 3)

The velocity of dolphin2 V_o = 12 mph

                                              = 5.33 m/s (Eq. 4)

Frequency of dolphin1 sound f₀ = 10000 Hz (Eq. 5)

Speed of sound in air V = 1531 m/s (Eq. 6)

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 10000*{[1531 + 5.33]/[1531 + 3.68]}

f’ = 10010.75 Hz

This is the frequency of sound as registered by the observer [dolphin2] under water when it is chasing the source [dolphin1] which is continuously moving away from the observer.

Difference in frequency = f’ – f₀

                                       = 10010.75 – 10000

                                       = 10.75 Hz

We notice that the difference is really small compared to the same scenario in air medium. It is because the source/observer velocity can be considered negligible with respect to speed of sound in water. However in air, the source/observer velocity is considerable with the speed of sound in air. Hence the difference in air would be 47.59 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.

Doppler shift underwater Pt7

DOPPLER SHIFT UNDERWATER PT7

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when one dolphin is chasing the other in a linear path. Consider two dolphins, dolphin1 [source] moving at 12 mph chasing dolphin2 [observer] which is moving at 8.3 mph. We’ll determine the apparent frequency as registered by dolphin2 when dolphin1 emits sound waves.

ASSUMPTIONS

1. Water has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of pressure is negligible

3. The amplitude of sound is unity

4. The water molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1) 

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in water at standard temperature and pressure conditions {V = 1531 m/s}

V_o – Velocity of dolphin2 [observer] (m/s)

V_s – Velocity of dolphin1 [source] (m/s)

The Doppler shift equation in this case is,

f’ = f₀*{[V – V_o]/[V – V_s]} (Eq. 2)

The ‘–’ sign in the numerator of equation (2) indicates that the observer is moving away from the source while the ‘–’ sign in the denominator indicates that the source is moving toward the observer.

  

The velocity of dolphin2 V_o = 8.3 mph

                                              = 3.68 m/s (Eq. 3)

The velocity of dolphin1 V_s = 12 mph

                                              = 5.33 m/s (Eq. 4)

Frequency of dolphin1 sound f₀ = 10000 Hz (Eq. 5)

Speed of sound in air V = 1531 m/s (Eq. 6)

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 10000*{[1531 – 3.68]/[1531 – 5.33]}

f’ = 10010.81 Hz

This is the frequency of sound as registered by the observer [dolphin2] under water when source [dolphin1] moves toward the observer.

Difference in frequency = f’ – f₀

                                       = 10010.81 – 10000

                                       = 10.81 Hz

We notice that the difference is really small compared to the same scenario in air medium. It is because the source/observer velocity can be considered negligible with respect to speed of sound in water. However in air, the source/observer velocity is considerable with the speed of sound in air. Hence the difference in air would be 48.86 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.