Star Wars Day: A Cobalt Conundrum

A Spoiler Warning: Though not extensive, I will be spoiling a small aspect of the movie for anyone who hasn’t seen it. It’s not a huge spoiler as such but still, spoilers ahead.

Having reconnected to the Star Wars universe in full, I was eager to commemorate Star Wars day properly. My knee jerk reaction was to embark on my journey of reviews of the   now numerous trades that house the thus far fantastic comic book stories that Marvel have masterfully crafted since 2015. However, after some reflection I realised that I could do something far more engaging (and, with hindsight, a great deal more challenging…).

Many of you will remember the opening scene of The Last Jedi (Whether or not that is a fond or bitter memory is entirely up to you). Following the successful evacuation of the Resistance base on D’Qar, Poe Dameron urges Leia to take action against the First Order’s Fulminatrix, a Mandator IV-class Siege Dreadnought capable of dealing extensive fire power against the Resistance’s lead vessel, the Raddus. Against her better judgment, Leia authorises the attack which is successful but at the cost of the entire MG-100 StarFortress fleet. Beyond the silver screen, confusion swept the audience (At least, I hope it did): How in the world could the bombs dropped onto the Fulminatrix “fall” down onto it, creating an impressive explosion, in space? Let’s find out!

Not an accurate representation by any means but it fits the purpose

Before we get started on the more exciting stuff, a bit about D’Qar. As detailed in the master piece above, D’Qar measures approximately 10,400 km (1). The Mass of planet is admittedly my own interpretation based on the relative size of Earth to D’Qar. If anyone knows the mass of D’Qar please let me know!

Question is, why is this important? Well, for the observant amongst you, you’ll have noticed that, whilst hardly close to the surface of D’Qar, the two ships were close enough for D’Qar to be seen in the frame. Admittedly, that doesn’t actually answer the question. The important thing is that they are close enough to the surface to at least experience some of its gravitational field.

Space Cricket is all the rage, so I’ve been told…

A gravitational field is a region around a body in which a force will experience a force due to the mass of the body. The picture above reflects the radial nature of the field around the planet, radial meaning that the further away you get from the body, the weaker the gravitational field is as a result of the distance between field lines doubling. It must  be stressed, though, that the field acts all around the body, not just where the lines are positioned. They merely represent the direction in which a body would move.

Newton’s Law of Gravitation states that the force due to gravity is proportional to the product of the masses of the bodies and inversely proportional to the square of their separation. That is a touch word but thankfully this transfers into a rather handy equation:


The negative sign indicates that the force is attractive which, when concerning gravity, it is always is (Luckily for us, otherwise we wouldn’t be able to walk anywhere). F stands for the Force due to Gravity, M is the mass of the large body, m the mass of the smaller body and r is the separation of the bodies. G is what we call the Universal Gravitational Constant which, as you might have guessed, is constant throughout the universe. It has a value of 6.67×10^-11 with the units Nm^2kg^-2.

If we relate this back to our conundrum we have just about enough data to work out the force felt due to gravity between the planet and one bomb. The trouble is, D’Qar isn’t the only thing that has a gravitational field around it…

An Artist’s impression of the Fulminatrix

I will confess that the value I’ve used for Mass is a somewhat of a guesstimate because apparently mass isn’t important for profiles on spaceships. I know, I’m as shocked as you are. For anyone who is interested I used the mass of a battleship and scaled it up by 1000. It’s smaller than D’Qar’s and that’s what matters. But wait, there’s more!

A marginally-more-successful Artist’s Impression of an MG-100 StarFortress

Though we are more focused on the bomb dropping down on to the Fulminatrix, one cannot simply dismiss the force, if notably smaller, between the bomb and its depositor which, again, is has a guesstimate mass based on that of a Hercules Jet (And again, what matters is that is is considerably smaller than both D’Qar and the Fulminatrix).

Now that we know the masses of all the main bodies in play (The bomb would experience forces from the rather large number of objects surrounding it but this could get out of hand very quickly). First things first, let us consider the forces of in each scenario.

IMG_0533 2
That should be 53.11N, my most sincere apologies

IMG_0534 2


From these calculations we can see that the D’Qar naturally has the greatest effect on the bomb relative to the Fulminatrix and the StarFortress simply because it is considerably larger than both of them.

From these values we can calculate the resultant force on the object using the equation:

F = F(D)+F(F)-F(S)


  • F – Resultant Force (N)
  • F(D) – Force of Gravity from D’Qar
  • F(F) – Force of Gravity from the Fulminatrix
  • F(S) – Force of Gravity from StarFortress

Which gives us a value of F=53.1N. In reality, this figure would increase as the bomb descends closer to the Fulminatrix and D’Qar and away from the StarFortress but, considering the small distance over which we are assuming it travels, the changes can be considered as negligible.

Now that we have the force experienced by the bomb we can use our good friend Newton’s Second Law of Motion which, in equation form, is F=ma to determine the acceleration of 1 bomb.

a= F/m

a= 53.1/5.5

a= 9.65 ms^-2

Now for some, that figure may look at least partially familiar. The acceleration due to gravity of Earth is equal to approximately 9.81ms^-2 (Though this is subject to tiny changes depending on where on the planet you are positioned). Part of the reason this value is so close is that A) their diameters aren’t that dissimilar and B) I used Earth’s mass to approximate the mass of D’Qar.

Using this value we can then determine the speed at which one bomb will strike the Fulminatrix and the time it takes to get there. Here we can call upon the SUVAT equations to help.

S – 30m   U – 0ms^-1   V – ?   A – 9.65ms^-2   T – ?

For V we will use: v^2 = u^2 + 2as which gives us a value of v = 24.1 ms^-1

For T we can use: t = (v – u)/a (A re-arrangement of v = u + at) which gives us a value of     t = 2.49s.

Question is, what does all this demonstrate? Well if we consider the original question, we’ve shown that Star Wars: The Last Jedi can just about get away with allowing bombs to “drop” (Though admittedly there is a whole lot of guesstimating with my values so take my “findings” with a degree of skepticism) when they are that close to a relatively large planet.

I hope you all enjoyed working through this little problem and that you’ve learned something about gravitational fields. If you have any thoughts or questions please feel free to leave them in the comments

May the Force the be with you!


  1. Wookieepedia (2018), D’Qar [online] Last Accessed: 4th May 2018, Available at:
  2. Lucasfilm Ltd. (2018) Resistance Bomber [online] Last Accessed: 3rd May 2018, Available at:

19 thoughts on “Star Wars Day: A Cobalt Conundrum

                    1. that’s literally what my dad did when he was younger, he rescued an injured jackdaw and it became his best friend – it used to follow my dad to school and attempt to get on the train with him (attack sparrows sound pretty fun too)

                      Liked by 2 people

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