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.