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Question: In his 1865 science novel From the Earth to the moon, Jules Verne described how three astronauts are...

In his 1865 science novel From the Earth to the moon, Jules Verne described how three astronauts are shot to the Moon by means of a huge gun. According to Verne, the aluminum capsule containing the astronauts is accelerated by ignition of nitrocellulose to a speed of 11 km/s along the gun barrel's length of 220 m.
(a) In g units what is the average acceleration of the capsule and astronauts in the gun barrel ?
(b) Is that acceleration tolerable or deadly to the astronauts ?
A modern version of such gun-launched spacecraft (although without passengers) has been proposed. In this modern version, called the SHARP (super High-Altitude Research Project) gun, ignition of methane and air shoves a piston along the gun's tube, compressing hydrogen gas that then launches a rocket. During this launch, the rocket moves 3.5 km and reaches a speed of 7.0 km/s. Once launched, the rocket can be fired to gain additional speed.
(c) In g units, what would be the average acceleration of the rocket within the launcher ?
(d) How much additional speed is needed (via the rocket engine) if the rocket is to orbit Earth at an altitude of 700 km ?

Explanation

Solution

We know that the gravitational force of the earth pulls everybody on the earth’s surface towards itself and the same goes for anything projected into the air. In order to escape anybody from the gravitational pull of the earth, that body has to be projected with escape velocity and its value on earth is 11.2km/s11.2km/s. Also, when some shot is fired using a gun, the bullet accelerates until the time it is inside the gun’s barrel and as soon as it comes out there is no acceleration in horizontal direction and there is only vertical acceleration, due to gravity, acting downwards.

Complete step by step answer:
(a) speed of the capsule= 11km/s=11×1000m/s=11000m/s11km/s= 11\times 1000m/s = 11000m/s.
Length of the barrel=220m220m
Initial velocity of the bullet is 0 because the bullet was at rest, so by using third equation of motion we get: v2u2=2as{{v}^{2}}-{{u}^{2}}=2as, where v is the final velocity, u is the initial velocity, a is the acceleration s is the distance covered.
v2u2=2as{{v}^{2}}-{{u}^{2}}=2as
v20=2as\Rightarrow {{v}^{2}}-0=2as
v2=2as\Rightarrow {{v}^{2}}=2as
(11000)2=2×a×220\Rightarrow {{(11000)}^{2}}=2\times a\times 220
a=275000m/s2\Rightarrow a=275000m/{{s}^{2}}
a=275000×ggm/s2\Rightarrow a=275000\times \dfrac{g}{g}m/{{s}^{2}}
a=275000×g9.8m/s2\Rightarrow a=275000\times \dfrac{g}{9.8}m/{{s}^{2}}
a=2.8×104g/s2\Rightarrow a=2.8\times {{10}^{4}}g/{{s}^{2}}
a=2272.72×g9.8m/s2\therefore a=2272.72\times \dfrac{g}{9.8}m/{{s}^{2}}

(b) The value is very high and for sure it will kill the passengers, so it is deadly.

(c) Now again using the third equation of motion to find the average acceleration inside the launcher.
v2u2=2as\Rightarrow {{v}^{2}}-{{u}^{2}}=2as
v20=2as\Rightarrow {{v}^{2}}-0=2as
v2=2as\Rightarrow {{v}^{2}}=2as
a=v22a\Rightarrow a=\dfrac{{{v}^{2}}}{2a}
a=(7000)22×3500\Rightarrow a=\dfrac{{{(7000)}^{2}}}{2\times 3500}
a=7000m/s2\Rightarrow a=7000m/{{s}^{2}}
a=7000×g9.8m/s2\Rightarrow a=7000\times \dfrac{g}{9.8}m/{{s}^{2}}
a=714g\therefore a=714g

(d) Now we want the rocket to orbit around the earth at an altitude of 700 km. So, make use of the law of conservation of energy.
Total height= R+h=7.07×106mR+h=7.07\times {{10}^{6}}m
mu22GMmR=mv22GMmr\dfrac{m{{u}^{2}}}{2}-\dfrac{GMm}{R}=\dfrac{m{{v}^{2}}}{2}-\dfrac{GMm}{r}
Here u is the initial velocity and v is the final velocity, R is the height final and r is the initial height.
Where, M=5.98×1024kgM=5.98\times {{10}^{24}}kg is the mass of the earth, substituting the value we get, the initial velocity is zero and so the first term gets cancelled out. Putting the value we find, v=6.05×103m/sv=6.05\times {{10}^{3}}m/s, but to orbit in this radius, v=GmMr=7.51×103m/sv'=\sqrt{\dfrac{GmM}{r}}=7.51\times {{10}^{3}}m/s
Finding the differences in the two speeds, vv=1.46×103m/sv'-v=1.46\times {{10}^{3}}m/s.

Note: While solving these types of problems we have to keep in mind that all the values are to be taken in standard SI units. Also, while solving problems involving planetary motion sometimes we make assumptions and often round the decimals in order to get some answer in a range. Sometimes we take value of acceleration due to gravity, g= 9.8m/s29.8m/{{s}^{2}} and sometimes we take the value of g to be 10m/s210m/{{s}^{2}}.