Gravity on Death Star 2

GRAVITY ON DEATH STAR 2

INTRODUCTION

The Death Star 2 is a battle station in the Star wars movie series. It was built by Imperial Military Research after the destruction of Death Star 1 by Luke Skywalker. The Death Star 2 makes its appearance in Star wars episode 6: Return of the Jedi. It is a battle station, meaning it can destroy almost anything using its super laser weapon. The fully functional super laser used here was more powerful than its predecessor. Although the laser was functional, the station was still under construction. The Death Star 2 is powered by larger Nuclear reactors right in its core. These reactors provide enough power to run the laser and the station as a whole. The Death star 2 is home to almost a million people including soldiers, droids, fighter jets, commanders, engineers and of course Darth Vader!

The construction of Death star 2 was carried out while it was orbiting planet Endor, which means it was a satellite. Eventually it was destroyed in the battle of Endor.

Let’s analyze some physical characteristics of the Death Star 2 like gravity, habitability and escape velocity.

ASSUMPTIONS

• The Death Star 2 is a homogeneous sphere
• It is completely made out of steel 

PHYSICAL CHARACTERISTICS

• The diameter of Death Star 2 [d] = 240 km
• So radius r = d/2 = 240/2 = 120 km = 120*10³ m
• The death star doesn’t rotate, so gravity will be same at pole and equator.

CALCULATION

Volume v = (4/3)*π*r³

                 = (4/3)*3.14*(120*10³)³

                 = 7.2345*10¹⁵ m³

Now Density ρ = mass/volume

Density of steel ρ = 7800 kg/m³

So mass   = density * volume

                = ρ * v

                = 7800 * 7.2345*10¹⁵

                = 5.6429*10¹⁹ kg

The acceleration due to gravity of Death Star 2

g = GM/R²

g – Gravity of Death Star 2

G – Universal Gravitation constant [G = 6.67 * 10^-11 Nm²/kg²]

M – Mass of Death Star [kg]

R – Radius of Death Star [m]

g = [6.67*10^-11 *5.6429*10¹⁹]/ [120*1000]²

   = 0.2613 m/s²

This value of g is smaller than that of moon but larger than that of its predecessor. It will not be easy to walk but very easy to jump as high as possible. The gravity will be same at both poles and the equator since the Death Star 2 also doesn’t rotate. With such a low gravity, one may feel weightless and move just by jumping. The gravity in fact is strong enough to prevent space workers from escaping to infinite space since it’ll always attract them. The Death Star 2 can overcome this situation by accelerating at 9.8 m/s² so that everybody on board feels just like earth.

However, since the station orbits planet Endor, it is always in a state of free fall like any other satellite. But it still has its own gravity and it will certainly play a role.

Now let’s determine the escape velocity of Death Star 2.

The escape velocity of Death Star 2

v = √2GM/R

   = √2*6.67*10^-11 *5.6429*10¹⁹/ [120*1000]

   = 250.46 m/s

This looks like a small number but it is not easy to achieve without any planes. One can escape the death star by using fighter jets but one cannot escape just by jumping. Even if one is successful in escaping Death star 2’s gravity, they’ll be subject to Endor’s gravity which they have to overcome in order to leave the system.

CONCLUSION

Thus we determined the acceleration due to gravity and escape velocity of Death Star 2. Both values were considerably small compared to Earth but sufficient enough to sustain and support life aboard.

Time dilation in Mount Everest

TIME DILATION IN MOUNT EVEREST

INTRODUCTION

The Earth performs circular motion once every 24 hours as it rotates about its own axis. Hence every object bound to the planet performs circular motion with the same angular velocity ω as that of Earth. Consider the tallest mountain in the world ‘Mount Everest’ which is 8848 m above sea level. This mountain also performs circular motion. Consider three observers one on ground level, one at the center of earth and another on the top of the mountain. The observer on the top performs a circular motion of greater radii with respect to the ground level in the same time which means the top observer is moving faster than ground level observer. The ground level observer also performs a circular motion relative to the observer in center of Earth. So according to the Special Theory of Relativity, a clock at the top of mountain would run slower than that on the ground level. We’ll find the time gained by top observer and ground level observer relative to the observer in the center of Earth.

ASSUMPTIONS

• Earth is a perfect homogenous sphere.
• The effect of Gravitational time dilation is negligible.

CALCULATION

The Angular velocity is given by,

ω = 2π/T [rad/s]

T – Rotational time period [T ≈ 24 hours = 86400s]

ω = 2*3.14/86400

ω = 7.2722*10^-5 rad/s

The angular velocity is same at all points on earth since it doesn’t exhibit differential rotation. But the tangential velocity on surface varies with the distance from the center. The greater the distance, higher is the velocity. It is important to note that since ground level observer is also performing a circular motion, he will also experience time dilation relative to observer at the center of Earth.

The tangential velocity of observer at center of Earth is given by,

v = [R]*ω [m/s]

R – Average radius of earth [R = 0 m]

v = 0 m/s

The tangential velocity of ground observer is given by,

v = [R]*ω [m/s]

R – Average radius of earth [R = 6371000m]

v = 6371000*7.2722*10^-5

v = 463.3118 m/s

The tangential velocity of mountain summit observer is given by,

v = [R + h]*ω [m/s]

R – Average radius of earth [R = 6371000m]

h – Height of the mountain [h = 8848 m]

v = 6379848*7.2722*10^-5

v = 463.9553 m/s

According to the Special Theory of Relativity, the time dilation equation is,

t' = t/γ [s]

t’– Mountain Top observer’s time. [s]

t₁ – Ground level observer’s time. [s]

t₂ - Proper time or center of earth observer’s time. [s]

γ – Relativistic gamma factor, γ = 1/√ [1-(v/c) ²]

c - Velocity of light [c = 3*10⁸ m/s]

t' = t*√ [1-(v/c) ²]

Center of Earth observer time

t₂ = t since v = 0

Ground level observer time

t₁ = t*√ [1-2.3850*10^-12]

t₁ = t*√ [0.999999999997615]

t₁ = t* 0.999999999998808

Mountain top observer time

t' = t*√ [1-2.3916*10^-12]

t' = t*√ [0.999999999997608]

t' = t* 0.999999999998804

CONCLUSION

We can observe that ground level and top observer’s time aren’t the same with respect to center of Earth observer’s time which proves that time dilates at a height. We’ll consider 5 different t₂ values and calculate t₁ and t' values. The larger the t₂ the more is the difference between t₁, t₂ and t’. Thus mountain top observer will gain some time over ground level and maximum time over center of earth observer. In fact mountain top observer will gain 37.7 microsecond in one year with respect to center of Earth observer.

Time	t2 [Center of Earth observer] (s)	t1 [Ground level observer] (s)	t' [Mountain Top observer] (s)
1 minute	60	59.9999999999285	59.9999999999282
1 hour	3600	3599.99999999571	3599.99999999569
1 day	86400	86399.999999897	86399.9999998967
1 month	2592000	2591999.99999691	2591999.9999969
1 year	31536000	31535999.9999624	31535999.9999623

Gravity on Death Star 1

GRAVITY ON DEATH STAR 1

INTRODUCTION

The Death Star is a battle station in the Star wars movie series. The first Death Star or Death Star 1 makes its appearance in Star wars episode 4: A New Hope. It took 20 years to build right from scratch. It is a battle station, meaning it can destroy almost anything using its super laser weapon. The Death Star is powered by Nuclear reactors right in its core. These reactors provide enough power to run the laser and the station as a whole. The Death star is home to almost a million people including soldiers, droids, fighter jets, commanders, engineers and of course Darth Vader!

The location of Death Star 1 is always classified and hence it is never close to any planet unless it’s planning to blow it up! This means the station is not a satellite but a free object in deep space. The Death star was eventually destroyed by Luke Skywalker by injecting a proton onto the heavily guarded reactor exhaust port at the northern hemisphere.

Let’s analyze some physical characteristics of the Death Star 1 and how will things turn out in the real world should there be a Death Star.

ASSUMPTIONS

• The Death Star is a homogeneous sphere
• It is completely made out of steel

PHYSICAL CHARACTERISTICS

• The diameter of Death Star 1 [d] = 160 km
• Radius r = d/2 = 160/2 = 80 km = 80*10³ m
• The death star doesn’t rotate, so gravity will be same at pole and equator.

CALCULATION

Volume v = (4/3)*π*r³                                                                                                                   (1)

                 = (4/3)*3.14*(80*10³)³

                 = 2.1435*10¹⁵ m³

Now Density ρ = mass/volume                                                                                                      (2)

Density of steel ρ = 7800 kg/m³

So mass = density * volume

                = ρ * v

                = 7800 * 2.1435*10¹⁵

                = 1.6719*10¹⁹ kg

The acceleration due to gravity of Death Star 1

g = GM/R²                                                                                                                                       (3)

g – Gravity of Death Star 1

G – Universal Gravitation constant [G = 6.67 * 10^-11 Nm²/kg²]

M – Mass of Death Star [kg]

R – Radius of Death Star [m]

g = [6.67*10^-11 *1.6719*10¹⁹]/ [80*1000]²

   = 0.1742 m/s²

This value of g is smaller than that of moon. It will not be easy to walk but very easy to jump as high as possible. The gravity will be same at both poles and the equator since the Death Star doesn’t rotate. With such a low gravity, one may feel weightless and move just by jumping although we didn’t see anything like that in the movie. The gravity in fact is strong enough to prevent space workers from escaping to infinite space since it’ll always attract them. The Death Star can overcome this situation by accelerating at 9.8 m/s² so that everybody on board feels just like earth.

Now let’s determine the escape velocity of Death Star 1.

The escape velocity of Death Star 1

v = √2GM/R                                                                                                                                  (4)

   = √2*6.67*10^-11 *1.6719*10¹⁹/ [80*1000]

   = 166.9698 m/s

This looks like a small number but it is not easy to achieve without any planes. One can escape the death star by using fighter jets but one cannot escape just by jumping.

CONCLUSION

Thus we determined the acceleration due to gravity and escape velocity of Death Star 1. Both values were considerably small compared to Earth but sufficient enough to sustain and support life aboard.

Time dilation in an airplane propeller

TIME DILATION IN AN AIRPLANE PROPELLER

INTRODUCTION

Airplane Propeller is the engine that powers the airplane. The diameter of Airplane Propeller varies from 5 feet [airbus models] to 12 feet [Boeing 777]. Consider a Jumbo Jet Boeing 777 Airplane Propeller which rotates at 481 mph and an observer who is stationary with respect to the Airplane Propeller. According to the Special Theory of Relativity, the Airplane Propeller’s clock would run slower compared to the observer’s clock. We’ll find the time gained by the Airplane Propeller relative to the stationary observer.

ASSUMPTIONS

Earth is a perfect homogenous sphere.

The effect of Gravitational time dilation is negligible.

CALCULATION

The Airplane propeller velocity is,

v = 481 mph = 213.88 m/s

According to the Special Theory of Relativity, the time dilation equation is,

t' = t/γ [s]

t’ – Actual time or Airplane Propeller’s time. [s]

t - Proper time or Stationary observer’s time. [s]

γ – Relativistic gamma factor, γ = 1/√ [1-(v/c) ²]

c - Velocity of light [c = 3*10⁸ m/s]

t' = t*√ [1-(v/c) ²]

t' = t*√ [1-5.0827*10^-13]

t' = t*√ [0.999999999999492]

t' = t * 0.999999999999746

CONCLUSION

We can observe that proper and actual time isn’t the same which proves that time dilates on Airplane Propeller relative to the stationary observer. We’ll consider 2 different t’ values and calculate t value. The larger the t’ the more is the difference between t and t’. Thus in one hour the Airplane Propeller gains 0.0152 nanosecond over the observer and in 15 hours it'll gain 13.7 nanosecond. So the more the propeller rotates the better ‘cause it’ll gain more time due to time dilation by rotating rather than just staying stationary.

Time	t’ [Stationary Propeller] (s)	t [Rotating Propeller] (s)	Difference (s)
1 minute	60	59.9999999999848	0.0000000000152
1 hour	3600	3599.99999999909	0.00000000091
15 hours	54000	53999.9999999863	0.0000000137