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.