Gravitational field of a Square plate

GRAVITATIONAL FIELD OF A SQUARE PLATE

INTRODUCTION

Gravity is derived from the Latin word ‘gravitas’ meaning mass. The universal law of gravitation was coined by Sir Isaac Newton. According to the law, any two masses anywhere in the universe separated by a distance will attract each other. This force of attraction is proportional to the product of their masses and inversely proportional to square of the distance between them.

(Eq. 1)

The distance between two masses can be finite or infinite, which is why gravitational force is referred to as long range force but is also the weakest force among all the other fundamental forces.

All objects that have mass will attract other masses. This means that each mass has its own gravitational field just like Earth. So this implies that all objects will attract each other since they will have their own field. This is not evident on Earth since Earth’s gravitational field outweighs all other mass’s field and hence all objects no matter how massive are attracted toward the Earth. In this post we intend to determine the gravitational field of a two dimensional Square plate and identify points where the plate’s own gravity is strong at some parts and weak at the other. 

ASSUMPTIONS

1. The thickness of the plate is negligible compared to its length and width.

2. The plate is not under the influence of an external gravitational field.

3. The plate is a homogeneous material.

4. All the mass is assumed to be concentrated at the center.

CALCULATION

Consider a square plate of length ‘a’ [m] and mass M [Kg]. We will first determine the center of gravity of this plate and then the magnitude of the gravitational field at points of interest.

Center of gravity

Fig 1 Square plate

The center of gravity of a regular square of side ‘a’ is (x,y) ≡ (a/2, a/2)

Points of interest and their distances from center

We will consider two points namely the corner and midpoint of a side.

Point A ≡ (0, a)

The distance between Point A and C.G. can be calculated by the distance formula

Point B ≡ (a/2, a)

The distance between Point B and C.G. can be calculated by the distance formula

Gravitational field

From Eq. 1, the force of attraction between a mass and its own surface is given by,                    

g – Acceleration due to gravity of the mass (m/s²)

G – Universal constant of gravitation

G = 6.67*10^-11 Nm²/Kg²

M – Mass of the object (Kg)

R – Distance between the centers of two masses (m)

G field at point A

The gravitational field beyond the surface is obtained by adding the additional distance,

d – Distance from the surface of the object to other object.

G field at point B

The gravitational field beyond the surface is obtained by adding the additional distance,

CONCLUSION

We thus determined the magnitude of gravitational field at various points on a square plate. We observe that gravitational field is stronger at point B than at point A, in fact it is twice as strong. This means that point B is attracted more toward the center because of its closer distance than point A which is slightly far away from the center.

Doppler shift for Bats

DOPPLER SHIFT FOR BATS

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 a Bat is hunting for food and is emitting ultrasonic waves in a linear path. The sound waves hit the prey and is reflected back which is detected by the Bat. So the prey is stationary but the sound waves reflected from the prey are moving toward the Bat. Consider a Bat and its prey, the bat is moving toward its prey at a speed of 99 mph [158.4 Kmph]. Initially the Bat is the source but later it is the observer since it’ll receive its own reflected sound from the prey which is the source. We’ll determine the apparent frequency as registered by the Bat when it hears its own reflected sound.

ASSUMPTIONS

1. The atmospheric air 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 humidity on sound is negligible

3. The amplitude of sound is unity

4. The air 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 air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

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

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

The Doppler Effect 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 toward the observer.

The velocity of source [Prey] V_s = 0 m/s {⸪ Source is stationary} (Eq. 3)

The velocity of observer [Bat] V_o = 99 mph

                                                       = 44 m/s (Eq. 4)

Frequency of Bat sound f₀ = 40000 Hz (Eq. 5)

Speed of sound in air V = 343 m/s (Eq. 6)

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

f’ = 40000*{[343 + 44]/[343]}

f’ = 45131.19 Hz

This is the frequency of sound as registered by the observer [Bat] when sound waves reflected from Prey approach the Bat.

Difference in frequency = f’ – f₀

                                       = 45131.19 – 40000

                                       = 5131.19 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the Bat due to Doppler shift. The Bat must consider this effect in order to catch its prey. At first it may not be aware of the Doppler shift hence may end up not accurately locating the prey. But with experience, it will be able to judge the position of prey based on the fact that emitted and received frequency aren’t the same.

Time lost by light in different mediums

TIME LOST BY LIGHT IN DIFFERENT MEDIUMS

INTRODUCTION

Light is an electromagnetic wave. It travels with a constant speed of 3 x 10⁸ m/s or 186000 miles/s in vacuum. It is also known as the cosmic constant c. But as the medium changes, the speed of light also changes. When light enters a medium, it undergoes refraction. Refraction is a phenomenon of bending of light. All mediums have a unique value of refractive index μ. Light still continues to move at c but it since it has to move through all the atoms of the corresponding medium, its speed is reduced by a factor of μ of the medium. All of the known mediums like water, glass, oil have μ greater than that of vacuum. For vacuum μ = 1. Thus light travels slowly in all mediums with respect to vacuum. We intend to determine the time lost by light in various mediums with respect to vacuum.

CALCULATION

The general equation for time lost by light is,

DT = (x/c)*[m - 1]

DT - Time lost by light (s)

x – Distance traveled by light (m)

c – Speed of light in vacuum (m/s)

MEDIUM 1: AIR [μ = 1.001]

The time lost in air is,

DT₁ = (x/c)*[1.001 - 1]                                                                  

       = 0.001[x/c] s

MEDIUM 2: WATER [μ = 1.33]

The time lost in water is,

DT₁ = (x/c)*[1.33 - 1]

       = 0.33[x/c] s

MEDIUM 3: GLASS [μ = 1.5]

The time lost in glass is,

DT₁ = (x/c)*[1.5 - 1]

       = 0.5[x/c] s

MEDIUM 4: GLASS, FLINT, 71% LEAD [μ = 1.8]

The time lost in flint glass is,

DT₁ = (x/c)*[1.8 - 1]

       = 0.8[x/c] s

MEDIUM 5: GLASS, ARSENIC TRISULFIDE [μ = 2]

The time lost in arsenic trisulfide glass is,

DT₁ = (x/c)*[2 - 1]

       = [x/c] s

Consider a beam of white light traveling a distance of x = 5 m through all mediums one at a time. We’ll determine the time lost by light in every medium and plot a graph of time lost versus medium to determine the medium where light loses maximum time.

Refractive Index [μ]	Time lost [s]
1.001	1.67E-11
1.33	5.50E-09
1.5	8.33E-09
1.8	1.33E-08
2	1.67E-08

  

GRAPH

EXPLANATION

We can observe that the graph is linear. It means that light will lose more time the as the refractive index increases. In this example light travels for 5 m in different mediums. When light travels through glass it loses around 8.3 nanosecond relative to vacuum and similarly 5.33 nanosecond in water. The loss will increase linearly with the distance traveled by light meaning if light were to travel for 1 year or 1 light year, the light in glass would lose considerable amount of time relative to the light in vacuum.

CONCLUSION

We thus determined the time lost by light in different mediums relative to vacuum.

Time lost by light in any medium w.r.t vacuum

TIME LOST BY LIGHT IN ANY MEDIUM WITH RESPECT TO VACUUM

INTRODUCTION

Light is an electromagnetic wave. It travels with a constant speed of 3 x 10⁸ m/s or 186000 miles/s in vacuum. It is also known as the cosmic constant c. But as the medium changes, the speed of light also changes. When light enters a medium, it undergoes refraction. Refraction is a phenomenon of bending of light. All mediums have a unique value of refractive index μ. Light still continues to move at c but it since it has to move through all the atoms of the corresponding medium, its speed is reduced by a factor of μ of the medium. All of the known mediums like water, glass, oil have μ greater than that of vacuum. For vacuum μ = 1. Thus light travels slowly in all mediums with respect to vacuum.

CALCULATION

Consider a light beam of any frequency traveling in vacuum. Let ‘x’ be the distance traveled by light, the time taken by light to travel this distance is,

t = x/c (Eq. 1)

t – Time taken by light (s)

x – Distance traveled by light (m)

c – Speed of light (m/s)

In any medium X, the speed of light will be,

c’ = c/m (Eq. 2)

c’ – Speed of light in medium (m/s)

m - Refractive index of medium

Time taken by light to travel the same distance x in medium X is,

t’ = mx/c (Eq. 3)

t’ – Time taken by light in medium (s)

If μ ˃ 1, then t’ ˃ t, thus time is always lost.

DT₁ = t’ – t

       = (mx)/c – x/c

       = (x/c)*[m - 1] (Eq. 4)

CONCLUSION

The above expression gives idea about time lost by light with respect to vacuum. The time loss is independent of frequency but depends on the speed of light. We can observe that time is always lost in all mediums of refractive index greater than 1.