a) Show that the rate of change of the free-fall acceleration with distance abov

Question

a) Show that the rate of change of the free-fall acceleration with distance above the Earth's surface is dg/dr = (? 2*G*ME)/RE^3 This rate of change with position is called a gradient. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is |?g| = (2*G*ME*h)/RE3 (Do this on paper. Your instructor may ask you to turn in this work.)

(c) Evaluate this difference for h = 28.50 m, a typical height for a ten-story building. m/s2

Explanation / Answer

g = GM/(R_E + r)^2

differentiate with respect to r

dg/dr = d/dr ( GM/(R_E + r)^2)

= GM d/dr ( 1/(R_E+r)^2)

= GM ( -2/( R_E+r)^3) d/dr ( R_E +r)

= - 2GM/( R_E+r)^3

if R_E >> r

dg/dr = - 2GM/R_E^3

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

for small differences

del g/ dt = del g/ h = 2G M_E/R_E^3

delg = 2GM_E h/R_E^3

(c)

delg= 2 x 6.67 x 10-11 x 5.97 x 1024 x 28.5 / (6371 x 103)3

= 8.77 * 10^-5m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.