Solutions to Special Relativity Problems (6)

We now examine the solutions from the previous introduction to special relativity instalment;

(1) A spaceship of mass 10^8 kg is to be accelerated to 0.6c using a matter-antimatter mix engine.

(a) How much energy does this require?

(b) How many kilograms of matter and antimatter will it take to provide this much energy?

Solutions

(a) by the work -energy theorem:

W = K(f) - K(i)

or:

W = m(o)c^2/ [(1 - u^2/c^2)^½ ] - m(o) c^2

where m(o) = 10^8 kg and u = 0.6c

Then:

W = (10^8 kg))c^2/ [(1 - (0.6c)^2/c^2)^½ ]

W = (10^8 kg)c^2/ ([1 - 0.36]^½) = (10^8 kg)c^2/ ([0.64]^½)

W = (10^8 kg)c^2/ (0.8) = 1.25 x 10^8 kg(c^2)= 1.12 x 10^25 J

(b) According to theory, equal amounts (masses) of matter and antimatter are required for complete annihilation and total conversion of the initial masses into energy. The energy needed as shown in (a) is 1.12 x 10^25 J. Thus: m(a) = m(m) = 6.25 x 10^7 kg.


(2) Consider the decay:

24 Cr 55 -> 25 Mn55 + e

The Cr 55 nucleus has a mass of 54.9279 u and the Mn 55 nucleus has a mass of 54.9244u.

(a) Calculate the mass difference between the two nuclei.

Solution:

delta m = 54.9279 u - 54.9244u = 0.0035 u


(b) What is the maximum kinetic energy of the emitted electrons?

The mass defect is thus 0.0035u and:

E = (delta m) c^2 =(0.0035u)(931 MeV/u) = 3.25 MeV

By the work -energy theorem:

W = K(f) - K(i)

K(i) = 0 (u1 = 0)

K(f) = m(o)c^2/ [(1 - u^2/c^2)^½ ] = 3.25 MeV

where we need to find u.

Using a table one finds the rest energy of the electron = 0.511 MeV

and K(f) = (3.25/ 0.511) = 6.36x the rest mass

So:

E = 6.36mc^2 = mc^2/ [(1 - u^2/c^2)^½ ]

6.36 = 1/ [(1 - u^2/c^2)^½ ] or:

(1 - u^2/c^2) = 1/(6.36)^2 = 1/40.44

or:

u^2/c^2 = 1 - 0.0024 = 0.9976

u = [0.9976c^2]^½ = 0.9987c



(3) Find the energy required to remove a single proton from 19 K 41.


In removing a single proton the atom remaining is: 18A 40

whose measured mass is: 39.96238 u

The mass of the final system becomes:

39.96238 u + 1.007825 u = 40. 970205 u

The effective mass increase of the system:

delta m = 40.970205 u - 40.96184 u = 0.008365 u

Then the energy needed to remove a single proton, from Einstein's eqn.

E = (delta m) c^2 = (0.008365 u) (931 MeV/u) = 7.79 MeV

(Which can also be converted to MJ using: 1 MJ = 1.6 x 10^-13 MJ/MeV)


(4)Find the speed and mass of an electron whose kinetic energy is 50 MeV.

Solution:

As per Problem #2, we know the electron rest mass = 0.511 MeV

Then, the mass of this electron is (50 MeV/0.511 MeV) = 97.84

or 97.84 times the normal electron rest mass.

The speed, following the same procedure applied in #2:

E = 97.84 mc^2 = mc^2/ [(1 - u^2/c^2)^½ ]

and we need u:

(1 - u^2/c^2) = 1/(97.84)^2 = 1/9572

or:

u^2/c^2 = 1 - 0.0001 = 0.9999

u = [0.9999c^2]^½ = 0.9999c (approx.)


(5) A rocket ship is to be accelerated to a speed of 0.5c. If propulsion is to be by using nuclear fuel, what fraction of the intial rest mass of the ship would have to be converted into kinetic energy to attain the desired speed?

What time dilation results if the speed is v = 0.5c?

Would this be sufficient to allow one generation of humans to reach the star Proxima Centauri (4.2 light years distant)?

Solution

by the work -energy theorem:

W = K(f) - K(i)

where K(i) is just the initial rest energy, or m(o)c^2

Then W = mc^2/ [(1 - u^2/c^2)^½ ] - m(o) c^2 = (total energy - rest energy)

Then total energy E:

= mc^2/ [(1 - u^2/c^2)^½ ] = mc^2/ [(1 - 0.25)^½ ]= 1.16mc^2

Then the excess (E - E' = E_k) is that required from the fuel, or:

E_k = (1.16 mc^2 - mc^2 = 0.16 mc^2)


Mas of nuclear fuel - call it m'- is then related to E_k by:

E_k = m'c^2 = 0.16m' c^2 or

(m'/m) = 0.16 or m' = 0.16m

In other words, the nuclear fuel mass(m') needs to be at least 16% of the total initial mass of the rocket. So, if the rocket's mass is 100,000 kg then the nuclear fuel must be at least (0.16)(100,000 kg) = 16, 000 kg.

For time dilation:

t' = t [1 - v^2/c^2]^½

again, v = 0.5 so:

t' = t [1 - 0.25]^½ = t [0.75]^½ = 0.866t

Now, there are 9.5 x 10^15 meters per light year

So the distance to Proxima Centauri is:

D = (4.2 Ly) x (9.5 x 10^15 m/Ly)= 4 x 10^16 m

Or: D = 4 x 10^13 km

The time required is then:

(0.866/0.500) x (4 x 10^13 km)/ (300,000 km/s]

= 1.73 x (4.22 yrs) = 7.3 years

Since a generation is generally figured as 40 years, the time factor will not be an issue.