Question

Luke Skywalker (m = 60 kg) enters the Death Star to save Princess Leia. Suppose the Death Star is composed of 101 floors of equal mass, M = 1 x10⁶ kg. Each floor is 10 m in thickness. Floor 0 is the outer surface. Luke enters at floor 1, beneath floor 0.

1. 1. How heavy is Luke when he is on floor 1? Assume there is no artificial gravity.
2. 2. Luke then enters an elevator and goes down to floor 80, where the detention block is located. How heavy is Luke at floor 80?
3. 3. If the floors had been equal density rather than equal mass, how would this have changed your answer?
4. 4. If you arrive at floor 100, at the very center of the Death Star, what would your weight be?

Answer 1

we have,

total number of floors

thickness of each floor

Since. Floor 0 is the outer surface

Radius of death star   

Let,

the number of arbitrary floor

So, the radial distance of floor ( bottom of the floor on which the Luke will be standing  ) will be given as,

_________________________________ relation 1

mass of each floor  

SO, the total mass enclosed within the sphere of radius

mass of the Luke

So, the gravitational force that is weight   of the Luke when he is in the floor will be given as,

  _________________________________ relation 2

Using and   in above relation we get,

_________________________________ relation 3

a)

So, whe the Luke is in floor 1

we have,

So, Using ,   and   in relation 3 we get the weight of the Luke in first floor as,

that is,

( Expressed in two significant figures )

b)

So, whe the Luke is in floor 80

we have,

So, Using ,   and   in relation 3 we get the weight of the Luke in first floor as,

that is,

( Expressed in two significant figures )

c)

Let,

density of every floor

we have

So,

total Volume elclosed below of floor that is volume of sphere of radius   

SO, the total ( new ) mass enclosed within the sphere of radius    

mass of the Luke

So, the gravitational force that is weight   of the Luke when he is in the floor will be given as,

Using relation 4 we get,

Using relation 1 in above relation we get,

So we see that

So we see that contrary to what we calculated in parts a and b we get,

d)

When the Luke arrives at center ,

we get, for first case of equal mass

and for second case of constant density,

So, using this in relation 2 we see that,

force at the very center of the Death star