Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Calculations

Notes to a video lecture on http://www.unizor.com



Rocket Calculation 1



Let's use the rocket equation

ΔV = −v_e·ln[m(t_beg)/m(t_end)]

to calculate how much propellant must be taken by a rocket to reach an orbit.



Here v_e is effective exhaust speed, m(t_beg) - mass of a rocket in the beginning of a time period during which a rocket's engine is working, m(t_end) - mass of a rocket at the end of this period of acceleration or deceleration.



Recall that the minus sign in this equation signifies the vector
character of the movement: positive direction of the exhausted
propellant (that is, the same as the movement of the rocket) causes
negative increment in rocket's speed - deceleration, while the negative
direction of exhausted propellant (that is, opposite to the movement of a
rocket) causes increase in rocket's speed - acceleration.



Contemporary rocket engine can have a very high effective exhaust
velocity. The speed of about 4km/sec is mentioned in a few sources we
are familiar with. So, we can assume that

v_e=4000m/sec.



An international Space Station's speed is about 7.8km/sec, as was calculated in one of the previous lectures on gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 7800m/sec



From this follows that

ln[m(t_beg)/m(t_end)] =

= 7800/4000 = 1.95



Therefore,

m(t_beg)/m(t_end) ≅ 7



So, the mass of a rocket at start is 7 times greater than its mass at
the end of its acceleration. For example, to launch 1,000 kg of useful
equipment and/or passengers to an International Space Station we will
need 6,000 kg of fuel.





Rocket Calculation 2



We still assume that

v_e=4000m/sec.



A rocket that goes far from Earth needs about 11.2km/sec speed to escape Earth gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 11200m/sec



From this follows that

ln[m(t_beg)/m(t_end)] =

= 11200/4000 = 2.8



Therefore,

m(t_beg)/m(t_end) ≅ 16



So, the mass of a rocket, that is supposed to leave the Earth's gravity,
at start is 16 times greater than its mass at the end of its
acceleration. For example, to launch 1,000 kg of useful equipment and/or
passengers to Mars we will need 15,000 kg of fuel.