The differential equation for a freee-falling mass m near the surface of the earth is given by: d2s/dt2=-g where s represents the distance from the surface of the earth to the object and the positive direction is considered to be upward. In other words, the underlying assumption here is that the distance s to the object is small when compared with the radius R of the earth; put yet another way, the distance y from the center of the earth to the object is approximately the same as R. If, on the other hand, the distance y to the object, such as a rocket, is large when compared to R, then we combine Newton's second law of motion and his universal law of gravitation to derive a differential equation in the variable y. Suppose a rocket is launched vertically upward from the ground. If the positive direction is upward and air resistance is ignored, then the differential equation of motion after fuel burnout is: m(d2y/dt2)=-k(Mm/y2) or d2y/dt2=-k(M/y2) where k is a constant of proportionality, y is the distance from the center of the earth to the rocket, M is the mass of the earth, and m is the mass of the rocket. To determine the constant, k, we use the fact that y=R, kMm/R2=mg or k[(gR2)/M]. Thus the last equation becomes: d2y/dt2=-g(R2/y2) Use the substitution v=dy/dt to solve the previous equation for v in terms of y. Assuming that the velocity of the rocket at burnout is v=v0 and y is approximately equal to R at that instant, show that the approximate value for the constant c of of integration is c=-gR+(1/2)v02. Use the solution for v in (1) to show that the escape velocity of the rocket is given by: v0=(2gR)1/2 The result in (2) holds for any body in the solar system. Use the values g=32ft/sec2 and R=4000 miles to show the escape velocity from the earth is approximately v0= 25,000 mi/hr. Find the escape velocity from the moon is the acceleration of gravity is 0.615g and R= 1080 mi.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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The differential equation for a freee-falling mass m near the surface of the earth is given by:

d²s/dt²=-g

where s represents the distance from the surface of the earth to the object and the positive direction is considered to be upward. In other words, the underlying assumption here is that the distance s to the object is small when compared with the radius R of the earth; put yet another way, the distance y from the center of the earth to the object is approximately the same as R. If, on the other hand, the distance y to the object, such as a rocket, is large when compared to R, then we combine Newton's second law of motion and his universal law of gravitation to derive a differential equation in the variable y.

Suppose a rocket is launched vertically upward from the ground. If the positive direction is upward and air resistance is ignored, then the differential equation of motion after fuel burnout is:

m(d²y/dt²)=-k(Mm/y²) or d²y/dt²=-k(M/y²)

where k is a constant of proportionality, y is the distance from the center of the earth to the rocket, M is the mass of the earth, and m is the mass of the rocket. To determine the constant, k, we use the fact that y=R, kMm/R²=mg or k[(gR²)/M]. Thus the last equation becomes:

d²y/dt²=-g(R²/y²)

Use the substitution v=dy/dt to solve the previous equation for v in terms of y. Assuming that the velocity of the rocket at burnout is v=v₀ and y is approximately equal to R at that instant, show that the approximate value for the constant c of of integration is c=-gR+(1/2)v₀²_.
Use the solution for v in (1) to show that the escape velocity of the rocket is given by: v₀=(2gR)^1/2
The result in (2) holds for any body in the solar system. Use the values g=32ft/sec² and R=4000 miles to show the escape velocity from the earth is approximately v₀= 25,000 mi/hr.
Find the escape velocity from the moon is the acceleration of gravity is 0.615g and R= 1080 mi.

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