# The weight of an object varies directly with its distance from the center of the Earth when the object is below the Earth's surface, but is inversely proportional to the square of its distance from the center when it is above the surface(Note that an object is weightless only because its weight is exactly counterbalanced by the upward force due to its motion)a)Phoebe's Small weighs 81 pounds at the Earth's surface. Write the particular equations expressing her weight as a function of distance form the cebter assuming the radius of the Earth is 4000 milesb)Predict Phoebe's weight 3000,8000, and 12,000 miles from the centerc)Draw a graph of this function in the domain from 0 through 12,000 miles

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by mrambabu1600
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2014-09-26T01:25:08+05:30

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Me = mass of Earth,  Re = radius of Earth,  G = universal Gravitational constant
d = distance between center of Earth and the object,  m = mass of an object.
C = GMm

Force of gravitation = weight = C / d²
on the surface or above the surface of Earth,   d >= Re.

Force of gravitation = K r  ,   where K is a constant
inside the surface of Earth.  Distance of object from center of Earth   r <= Re.

On the surface  weight = 81 pounds.   Re = 4000 miles

Force at the surface according to both formulas is same.
C / Re² = K Re
So K =  C / Re³

The force of gravitation below surface of Earth = weight = K r = (C / Re³ ) r
= [ C/Re² ] * r / Re
= 81 pounds * r / 4000 miles
= 0.02025 r   units
Weight = (0.02025 * r) pounds,  r distance of phoebe from center in miles.
(weight has to be multiplied by g in FPS units to get the value in units of force.)

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b)
Weight at 3000 miles is = 0.02025 * 3000 = 60.75 pounds

weight at surface = C / Re² = 81 pounds   =>          C = 81 * 4000²  units

Weight at 8000 miles = C / 8000² = 81 * 4000²/8000² = 20.25 pounds

Weight at 12000 miles = C / 12000² = 81 * 4000² /12000² = 9 pounds

C )  see diagram