Chapter 7 Models for Wave and Oscillations

Variable Gravitational acceleration– Lunar Lander– Escape velocity

Wave motion and interaction– Pendulum without damping– Pendulum with damping

Mechanical variation

Variable Gravitational acceleration

Motion with constant gravitational acceleration: motion remains in the immediate vicinity of the earth’s surface– Horizontal motion: Train, MRT, walk & run, ship, …..– Vertical motion with lower speed: Airplane, jump from an airplane, bullet, …..

• Radius of the earth: 6378(Km) & height of airplane: 10 (km)

Motion with variable Gravitational acceleration: a projectile in vertical motion doesn’t remains in the immediate vicinity of the earth’s surface – Rocket, Space-shuttle– Missile– Apollo lunar lander– Satellite, ……

Variable Gravitational acceleration

Problem: The displacement traveled in vertical direction is comparable with the radius of the earth.

Newton’s law of gravitation: The gravitational force of attraction between two point masses M and m located at distance r apart is given by

– G is a certain empirical constant – The formula is also valid if either or both of the two masses are

homogeneous spheres; in this case, the distance r is measured between the center of the spheres.

2

G M mF

r

11 26.6726 10 ( / )G N m kg

Example: A lunar lander

The problem: A lunar lander is free-falling toward the moon’s surface at a speed of v0=450 m/s (that is, 1620 km/h). Its retrorockets, when fired in free space, provided a deceleration of T=4 m/s2. At what height above the lunar surface should the retrorockets be activated to ensure a ``soft touchdown’’ (v=0 at impact)?

– Mass of the moon: M=7.35E22 (kg)– Radius of the moon: R = 1.74E6 (m)

Solution: – let r(t) denote the lander’s distance from the center of the moon at time t– When we combine the (positive) thrust acceleration T and the (negative) lunar acceleration F/m, we get

Example: A lunar lander

The second order ODE:

The problem:

Re-write in terms of the lander’s velocity:

Substitution using the chain rule formula

2

2 2

( )total force

( )

d r t G Mm a mT F T

dt r t

1 1 1Find (0) such that at , ( ) & ( ) 0r t t r t R v t

2

dr dv G Mv T

dt dt r

2: Here as new independent variable!!

dv dv dr dv dv G Mv v T r

dt dr dt dr dr r

Example: A lunar lander

Integration of both sides with respect to r

From the ``soft touchdown’’ condition

The solution:

Solution: – Find r when v=-450m/s with – T=4m/s2, G=6.6726E-11, M=7.35E22, R=1.74E6

21

2

G Mv T r C

r

0G M

r R v C T RR

2 2 21 1( ) 0

2 2

G M G M G Mv T r T R T r T R v r G M

r R R

Example: A lunar lander

The equation for r:

The two roots:

The result: The lander’s desired initial height above the lunar surface is

2 6 124 (9.87985 10 ) 4.90436 10 0r r

6 60.68809 10 (drop since <R) 1.78187 10 (accept)r r

1781870 1740000 41870( ) 41.87( )r R m km

Escape velocity

In 1865, Jules Verne (in his novel from the earth to the moon) raised the question: – What is the initial velocity necessary for a projectile fired from the surface of the earth to reach the moon?? – A similar question: What is the initial velocity necessary for the projectile to escape from the earth altogether??

Condition: If the velocity v remains positive for all t>0,

so it continuous forever to move away from the earth!!

Variables: – r(t): the projectile’s distance from the earth’s center at time t

( )( ) : velocity

dr tv t

dt

Escape velocity

The equation

– Parameters:• Mass of the earth: M=5.975E24 (kg)• Radius of the earth: R=6.387E6 (m)• The constant: G=6.6726E-11 N(m/kg)2

Substitution using the chain rule formula

2

2 2

( )

( )

dv d r t G M

dt dt r t

2: here r as the new variable!!

dv dv dr dv dv G Mv v

dt dr dt dr dr r

Escape velocity

Integration of both sides with respect to r

Initial condition

The constant C:

The solution:

21

2

G Mv C

r

00 : &t r R v v

2 20 0

1 1

2 2

G M G Mv C C v

R R

2 20

1 12 ( )v v G M

r R

Escape velocity

The velocity is positive for all t>0 or r>R:

The escape velocity:

The escape velocity from the earth

2 2 2 20 0 0

2 2 2 2G M G M G M G Mv v v v

R r R R

0

2G Mv

R

11 24

0 6

2 2 6.6726 10 5.975 10

6.378 10 11180 / 11( / ) 40000( / )

G Mv

Rm s km s km h

Escape velocity

The escape velocity from the moon

It is just over one-fifth of the escape velocity from the earth’s surface!!!The escape velocity:– The heavier the sphere, the larger the velocity– The thicker the sphere, the smaller the velocity

11 22

0 6

2 2 6.6726 10 7.35 10

1.74 10 2375 / 2.4( / ) 9000( / )

G Mv

Rm s km s km h

Models for pendulum motion

Without damping

With damping