If a ball were dropped in a hole bored through the center of the earth, it would

Question

If a ball were dropped in a hole bored through the center of the earth, it would be attracted toward the center with a force directly proportional to the distance of the body from the center. Find: the frequency of the motion.

Note:

when the body (mass = m) is on the surface of the earth,

the attraction force is F = GMm/R2, where M is the mass of the earth, and R is the radius of the earth.

When the body is "below" the surface, say, with distance y from the center.

You may think the body is on the surface of the "sub-earth" whose radius is y.

And the mass of the "sub-earth" is Ms = M (y/R)3.

So the attraction force now is F = GMsm/y2 = (GMm/R3)y which is proportional to y.

To do this problem, you will need to use numbers R=4000 miles and GM/R2 = g = 32 ft/sec2

Explanation / Answer

Mass of the ball = m

Mass of Earth = M

We will convert all given units (ft/s and miles) to SI units for our calculations

Earth Radius R = 4000 miles = 6437.4 km = 6.45 * 106 m

g = 32 ft/s2 = 9.75 m/s2

Force on the ball when it is on the surface of earth: F = GMm/R2

Force on ball due to earth's gravitational field = F = GMs m/y2

Where y = distance of ball from centre of earth

Ms = Mass of sphere with radius r' when the ball is below the surface of earth.

When ball is on surface of earth, y = R, Ms = M

Mass of sphere 'sub-earth': Ms = M (y/R)3

Thus, we can write F = GMs m/y2 = (G Mm/R3 ) * y

Now from Newton's second law of motion, force is also written as:

F' = ma = m * d2y/dt2

To find equation of motion of the ball, we equate the two forces above:

F' = F (however the direction of two forces are opposite)

So, F' = -F

or,

Now since the force of gravitational field is directed towards the centre of Earth (opposite to the displacement from the centre), we have put a -ve sign to take care of this in the above equation:

m * d2y/dt2 = -(G Mm/R3 ) * y

or,

d2y/dt2 = -(G M/R3 ) * y    ................................................(1)

Now when ball is on the surface of earth,

We use it to find expression for g:

F = ma = mg = GM m /R2

or, g = GM/R2

Substitute this in (1) above:

d2y/dt2 = -(G M/R3 ) * y = -(GM/R2 ) * y/R = -(g/R) * y

Putting the given values of g and R, we have:

d2y/dt2 = -(9.75/6.45*103) * y = -1.5 * 10-6 y

This is the required equation of motion of the ball.

We can write this as:

d2y/dt2 = -1.5 * 10-6 y

or,

d2y/dt2 + 1.5 * 10-6 y = 0

Now compare this equation with standard equation of motion for simple harmonic motion:

d2y/dt2 + w2 y = 0

where, w = angular frequency of oscillation

w2 = 1.5 * 10-6   /s2

or,

w = square root (1.5 * 10-6) = 1.23 * 10-3 oscillations/s

or,

w = 1.2 * 10-3 * 60 oscillations per minute = 72 * 10-3 = 0.07 oscillations per minute = 0.07 * 60 = 4.2 oscillations per hour.

To calculate linear frequency of oscillation, we know: w = 2*pi*v

v = linear frequency

Thus, linear frequency: v = w/2*pi = 4.2 /(2*3.14) = 0.67 = 0.7 oscillations per hour.............ANSWER

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