12. A m = 71.2 kg object is released from rest at a distance h = 0.713515 R abov
ID: 2032111 • Letter: 1
Question
12. A m = 71.2 kg object is released from rest at a distance h = 0.713515 R above the Earth’s surface. The acceleration of gravity is 9.8 m/s 2 . For the Earth, RE = 6.38 × 106 m, M = 5.98 × 1024 kg. The gravitational acceleration at the surface of the earth is g = 9.8 m/s 2 . Find the speed of the object when it strikes the Earth’s surface. Neglect any atmospheric friction. Caution: You must take into account that the gravitational acceleration depends on distance between the object and the center of the earth. Answer in units of m/s.
Explanation / Answer
we will use conservation of energy and gravity to find the speed
h = 0.713515 R = 0.713515 ( 6.38 × 10^6) = 4.5522 x 10^6 m
g ( above the surface) = GMe/ ( h+ R)^2= 6.67×10^-11. x 5.98 × 10^24 / (
GMm/ ( h + Re) = 1/2 mv^2-------------------eq(1) ( v is the velocity with which the object enters the surface of earth)
GM / ( h + Re) = 0.5 v^2
6.67 x 10^-11 ( 5.98 × 10^24) / ( 4.5522 x 10^6 + 6.38 × 10^6) = 0.5 v^2
79.7732 x 10^ 7 / ( 10.9322) = v^2
v = 8.54 x 10^ 3m/s
V ( speed with wiich the object touches the surface of earth) = sqroot ( v^2 + 2 x9.8 x6.38 × 10^6)
V = sqroot (79.97 x 10^ 6 + 2 x9.8 x6.38 × 10^6) = sqroot (205.018 x 10^ 6) =14.318 x 10^ 3 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.