Question

1. The space force developed a new weapon system that makes an object experience no air drag. The objects have a mass of 100 kg. If one of the objects is dropped from a space weapons platform from a height of 136.3 miles, how much Kinetic Energy would it be able to release upon impact? ( Only work in the y dimension) A stick of dynamite has 1.7 MJ of energy. How many sticks of dynamite equivalent would this be?

2. It takes a delta launch vehicle 8 min to get to low earth orbit about 100 miles up. If it went in a straight line (it doesn’t) and its speed on orbit was 17,500 mph, how many g’s would the space force service members go through? Ignore the rotation of the Earth and just solve as the rocket is in one dimension. Convert all values to metric and show the conversions.

Answer 1

1) The energy of the weapon will be the potential energy it has when it is at this height.

The height from which the weapon is falling = 136.3 miles = 136.3*1609 = 219353.59 m

The mass of the weapon = 100 kg

So, The potential energy is

U = mgh = 100*9.8*219353.59 = 214.967 *10^6 J = 214.967 MJ

A stick of dynamite = 1.7 MJ

So, This energy is equal to 214.967/1.7 = 126.45 dynamite sticks.

2) The satellite is at a height of

H = 100 miles = 100*1609 = 160900 m

The potential energy of the vehicle is

U = mgh = m*9.8*160900 = 1576820 J

The speed of the vehicle at this height is

v = 17500 mph = 17500/2.237 = 7822.98 m/s

So,

Kinetic energy is

K = m*7822.98^2/2 = m*30599508 J

SO, total energy is

E = U + K = m*(30599508 + 1576820) = m*32176328 J

So, the kinetic energy at the bottom will be equal to this

mv^2/2 =m* 32176328

v = sqrt(2*32176328) = 8022.01 m/s

The time taken = 8 min = 8*60 = 480 s

So,

acceleration is

a = (v-u)/t = (8022.01)/480 =16.71 m/s^2

In terms of g, this is

a = 16.71/9.8 = 1.7 g