Solutions to Rotational Dynamics Problems (4)

1) Consider the table shown below, applied to circular satellite orbits around the Earth:

Distance from Center of Earth	2 r	3 r	3r/ 2	4 r	5 r/2	5 r
Acceleration of gravity	g/4

 Complete the Table above given that r = 6.4 x 10 ⁶ m. Hence or otherwise, deduce the acceleration of gravity g at a distance 10r.

Solution:

The solution here is based on noting that the gravitational field intensity g ~ 1/r²

That is, it is inversely proportional to the distance from the center squared. Hence, the remaining entries in the table as follows, in ascending order:

 g/9, 4g/9, g/ /16, 4g/25   and g/25

Given this relationship (g ~ 1/r²) then at a distance 10r we have: g/100,
Thus, if r = 6.4 x 10⁶ m, then at a distance:

 10 r = 10 (6.4 x 10⁶ m) = 64 x 10⁶ m = 6.4 x 10⁷ m

And given, g = 9.8 ms^-2, we would find a new gravitational intensity  (corresponding to the much greater distance from the center) of:

g/100  =   (9.8 ms^-2)/ 100 = 0.098 ms^-2

2) Newton’s version of Kepler’s 3^rd law is usually written as:

 (m1 + m2)P² = 4p²/G (r1 + r2)² 

where G is the Newtonian gravitational constant: 

(G = 6.7 x 10^-11 N-m²/kg²) and (r1 + r2) is the distance between the centers of the two bodies.

 Use this to find the period P of the Moon if its mass m1 =  m2/80 where Earth’s mass m2 = 6.0 x 10 ²⁴ kg and (r1 + r2) = 384, 000  km.    Compare this to the value obtained by using Kepler’s simpler version of the 3^rd law with (r1 + r2) =  a1= 0.0025 AU.

Soln.

m2/80 = (6.0 x 10 ²⁴ kg)/ 80 = 8.3 x 10 ²² kg

(r1 + r2) = 3.84 x 10 ⁵  m

P = Ö4p²/G (a1)³  (m1 + m2)

G =  6.7 x 10 ^-11   m ³  s^-2 kg ^-1

(m1 + m2) = 6.0 x 10 ²⁴ kg  +  8.3 x 10 ²² kg » 6.0 x 10 ²⁴ kg

P =

 Ö 4p²/ (6.7 x 10 ^-11   m ³  s^-2 kg ^-1) (0.0025 AU ) ³  6.0 x 10 ²⁴   kg

P »  0.002 yr.

3) Verify the conservation of angular momentum applies for a spacecraft in orbit around the Earth if its velocity at perigee is  10.7 km/ sec, its distance from Earth at perigee is  6.6 x 10 ³ km, its velocity at apogee is  0.75  km/ sec  and its distance at apogee is: 9.3 x 10 ⁴ km.  Find the period of the spacecraft.

Soln.

For conservation of angular momentum L in an elliptical orbit (with r_a  the radius vector at aphelion,  and  with r_p the radius vector at perihelion)

 L = mv_a r_a =  mv_p r_p

Or: v_a r_a =  v_p r_p

v_a r_a =    (0.75  km/ sec) (9.3 x 10 ⁴ km) » 7.0 x 10 ⁴ km²/sec

v_p r_p =    (10.7 km/ sec) (6.6 x 10 ³ km) » 7.0 x 10 ⁴ km²/sec

Given:  v_a r_a »  v_p r_p

To within the limits of errors, the principle of angular momentum conservation  is validated.

The period is obtained from: T = 2π (a³/m)^½

Where m  = G (m1 + m2) 

And:  a = [r_a +  r_p ] /2 =  (9.3 x 10 ⁴ km + 6.6 x 10 ³ km)/2 »

= (9.3 x 10 ⁷ m + 6.6 x 10 ⁶ m)/2 » 4.9 x 10 ⁷ m

T »

2π [(4.9 x 10⁷ m)³/[6.7 x 10^-11 Nm²/kg²) (6.0 x 10²⁴ kg)]^½

T » 4.3 x 10 ⁴ s » 12.2 hrs.

4) A gyroscope is made from a flywheel made to spin about a rotation axis (bar) as shown in the sample diagrams in the text. If each of the two forces applied = 20 N and the distance d is 0.15m, then find the magnitude of the torque t.

If the mass of the flywheel is 1.0 kg and its radius r = 0.1 m find the magnitude of the angular momentum if  w =  p rad/sec. Thence or otherwise find the rate of precession of the angular velocity W.

Soln.

t    =   2F d = 2 (20 N) 0.15m =  6 Nm

The angular momentum:

L = Iw =  (½ m r ² ) ( p rad/sec) =

½ (1.0 kg) (0.1m) ²  ( p rad/sec)  =    0.016 kg m² /sec

The rate of precession:

 W   = 2F d/ Iw  =   6 Nm/ 0.016 kg m² /sec =  375 kg m² /sec

5) (Inquiry problem)

A proponent of Immanuel Velikovsky's "colliding worlds" nonsense claims that all that was needed for Earth's rotation to stop (when Venus came so close to Earth in passing from Jupiter to its present orbit) was a torque defined:  N = F_r X R_c.  as used in his diagram below:

The problem is that an OFF-axis torque  (e.g. perpendicular force ) is needed to effect this, and the only "torque" he has is mythical, based on: N = F_r X R_c. But F_r is a fantasy force that doesn't exist! The only actual force acting is an attractive (gravitational) one through the centers! Hence no external agent is available to act perpendicular to the line of centers to halt the rotation! 

6) Inquiry Problem

In the early morning hours of June 14, 2002, the Earth had a  remarkably close encounter with an asteroid (2002 MN) the size of a small city. It remained undetected until three days after it had passed the Earth. At its closest approach, the asteroid was 73,600 miles from the center of the Earth—about a third of the distance to the Moon. Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it had an average density of 3.3 g/ cm ³

------------------

Hint:  Use  E _tot = K + V  = mv² / 2 -   GMm/r =  -   GMm/2r

And take  the asteroid as coming from 'infinity' with r the distance to Earth at closest approach.

Soln.

E _tot = K + V  = mv² / 2 -   GMm/r =  -   GMm/2r

To obtain velocity:

E _tot  = mv² / 2 -   GMm/r =  -   GMm/2r

And: v² / 2 -   GM/r =  -   GM/2r

v² / 2  =  -   GM/2r + GM/r =   GM/2r

Then velocity: v = Ö GM/2r

Where r =   73,600 miles =  117, 600 km = 1.17  x 10 ⁶ m

We know M (Earth's mass) = 6.0 x 10 ²⁴   kg

G =  6.7 x 10 ^-11   m ³  s^-2 kg ^-1

Average density of asteroid given as  3.3 g/ cm ³  = 3300 kg/ m ³ 

Mass of asteroid = density x volume

Assume spherical volume so V = 4/3  (p  r ³ )

Where r = 2.0 km =  2.0  x 10 ³ m

The Volume V = 4/3  (p (2.0  x 10 ³ m)³ ) = 3.3  x 10 ¹⁰ m ³ 

Mass of asteroid = (3300 kg/ m ³   ) 3.3  x 10 ¹⁰ m ³ 

= 1.1 x 10 ¹⁴  kg

v = Ö GM/2r =  where M = 6.0 x 10 ²⁴   kg

G =  6.7 x 10 ^-11   m ³  s^-2 kg ^-1

v = Ö (6.7 x 10 ^-11   m ³  s^-2 kg ^-1)  6.0 x 10 ²⁴   kg/ 2(1.17  x 10 ⁶ m)

= 1.3 x 10 ⁴ m/ s

The kinetic energy of the asteroid is:

K = 9.6  x 10 ²¹ J

 For reference, this magnitude approaches the energy of the 

Chicxulub asteroid at  300 zetta joules.