사이언스 올제

읽고 생각하는 과학문제

● 10월호 문제 풀이 ☞10월호 문제

1. The Orion launch-abort system can lift the crew module with a thrust 15 times its weight. How long a "burn" is needed to lift the crew 300m vertically? Answer is about 2.0 s.

The force developed by the abort rocket engines is F = 15mg, and the module must be lifted vertically out of the way of the Orion launch rocket to a vertical height of 300m. The acceleration of the module, which we assume to be constant, can be found from "F=ma" and since the net force on the module includes gravity pointing down,

"F=ma" is

+15mg - mg = ma

and the acceleration is a=14g, and the acceleration of gravity is g=10 m/s^2 since we are still close to the surface of Earth. The distance traveled under acceleration a is d = 1/2 a t^2, so the time required is

t = √( 2 d / a ) = √( 600m / 14g ) =

√(600/140) = 2.07 s.

This is a very fast ride! In fact, most people will "black out", or faint, under this extreme acceleration.

2. The booster Ares I attains a speed of Mach 6 in two and one-half minutes. Mach 1 is 340 m/s (the speed of sound in air). What is its acceleration? The acceleration is about 14 m/s^2.

This problem is the relationship between velocity and acceleration,

and the velocity after a time t is

v = a t,

so the acceleration a is

a = v / t = Mach 6 / 2.5 minutes = 6 x 340 m/s / 150 s = 1.36 m/s^2,

or about 1.36g of acceleration.

3. The Earth return module has mass M, velocity v, and enters the atmosphere at an altitude of h = 200 km. How much heat energy is generated before landing? Answer is 1/2 M v^2 + Mgh.

Getting into space is easy; it just requires a big enough rocket and the Chinese have had rockets for 1000 years. The problem is getting back. In orbit, a module of mass M has a total energy that is kinetic energy (KE) and gravitational potential energy (GPE).

GPE is approximately Mgh, where h is the height (we are just approximating g as constant over the height h, although it becomes smaller as you move away from Earth).

The KE is 1/2 M v^2, a number which is huge since v ~ 30,000 km/h. In order to land safely, the module must lose all this energy, and the only mechanism is friction with the air of the atmosphere. There is simply no other force available. (There are small forces involving electromagnetic induction from Earth's magnetic field, but this is too small).

This friction can only appear as heat which must be radiated away from the module, and the total energy to be radiated is 1/2 M v^2 + Mgh.

4. The escape velocity from Earth is 11.2 km/s. If the Moon's mass is 0.0123 that of Earth's mass, and the Moon's radius is 0.273 that of Earth's, what is the escape velocity from the Moon?

The escape velocity is 2.4 km/s.

In order to "escape" from a mass M, you must overcome the gravitational force of attraction and be able to reach "infinity". An easy way to see this is to consider energy. When a mass m is in the gravitational field of a mass M, its gravitational potential energy is

GPE = - G M m / r.

where r is the distance between masses and G is the gravitational constant. This is the integral of the force from r out to infinity and is the conventional form in which GPE becomes zero at infinite separation. Notice the negative sign: a mass m is "confined" to the potential. In order to escape,

the mass m needs a KE large enough to bring its total energy up to zero, or slightly above zero, that is, KE + GPE = 0. This is

1/2 m v^2 - G M m / r = 0,

or a velocity of

v = √( 2 G M / R )

when starting from the surface of Earth at radius R. We do not know M and R in the problem, and we don't even need to know G. We can do a favorite thing that physicists love to do: we can "scale" from a known solution to this solution. The escape velocity scales like

√( M / R ),

that is, v_escape is bigger by √M and smaller by √R.

So,

v_escape_Moon = v_escape_Earth * √( M_Moon / M_Earth ) *

√( R_Earth / R_Moon).

This is,

v_escape_Moon = (11.2 km/s) * √(0.0123) * √( 1 / 0.273 ) =

2.38 km/s.

★ 참고사항 : √= sqrt