example Real-World Engineering Problem

What you need to know:

v = at + v0 where a is acceleration and v0 is initial velocity

Earth’s gravitational force acts on any object in the proximity of earth. This includes satellites in low earth orbit, satellites in geostationary orbit and the moon. The force is always directed toward the center of the earth.

Solve the dynamics problem, below:

A rocket launched into Low Earth Orbit (LEO) accelerates from launch until the thrust stage is complete. As thrust continues, the mass of the rocket decreases. If the thrust remains constant, the acceleration of the rocket will itself increase. Therefore, in order to determine the acceleration of a rocket, the weight of the fuel as well as the rate at which fuel is burned must be known.  

a. Let's start with the assumption that the acceleration of the rocket is constant. In order to reach low earth orbit (LEO), the rocket must reach the speed given by the equation for which the force needed to keep the rocket going in a circle (as opposed to flying straight) is equal to the gravitational force on the rocket. We write mv^2/r, the "centrifugal" force," is equal to the force of gravitational attraction of any two bodies. That force is given by Gm1m2/r^2, where m1 and m2are the masses of the two bodies, r is the distance between the center of mass of each body, and G is known as the "Gravitational Constant", and has a value of 6.7E-11 in MKS units. What is the speed required to achieve LEO?

b. v = at + v0 where a is acceleration and v0 is initial velocity. Let's see if we can figure out the acceleration of the rocket as a function of time, assuming we know the rate of fuel burn. We find that Elon Musk's Falcon Heavy can hold about 200,000 kgs of fuel and it burns at a rate of 1500 kg per second. At that burn rate, the thrust is 1.5E7 Newtons. Assuming the burn rate is constant and the initial weight of the rocket is 1.4E6 kg, how fast is it going after burning all its fuel? Answer in km/s.

Answers to Problem I

a. 28,000 km/hr

b. 16 km/s

Guide to Problem I Answers

a. "m" in the term mv^2/r, is equal to "m1" in the term Gm1m2/r^2, so equating the two terms gives Gm2/r = v^2, where one of the r's cancelled. m2 is the mass of the earth, which is 6E24 kg. The radius of the earth is 6.4E3 km. Solving for v gives v = √(6.7E-11 x 6.0E24 / 6.4E6 = √ 6.3E7 = 2.5E5. So low earth orbit speed is 7.9E3 m/s, or 28,000 km/hr.

b. We know that Newton’s 2nd law, F = ma is always true. Since F is constant, we need to find equations which describe m and a as a function of time. The mass of the ship is going to be its initial mass minus the spent fuel. We can write m(t) = mi - rbt, where m(t) is the mass as a function of time, mi is the initial mass and rb is the fuel burn rate. Then a = F/(mi - rbt), where F, mi and rb are constant and known. 

The velocity v is given by at + vo. Since vo is zero, we just need to integrate the acceleration from time t = 0 to time to t = 135 sec. 

Then v = Ft/(mi – rbt)

At this point we can just plug in values to get:

a = 1.5E7 / (1.4E6 - 1.5E3 x t).

The time to burn all the fuel is 2E5/1.5E3 = 135 seconds. 

v = 120 x 135 = 16,000 m/s, or 36,000 mph.