A projectile is shot directly away from Earth\'s surface. Neglect the rotation o

Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.

(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is 0.590 of the escape speed from Earth?
RE

(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is 0.330 of the kinetic energy required to escape Earth?
RE

(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
J

Explanation / Answer

escape speed on earth VE = sqrt(2GM/RE)

(a)

if projectile speed v = 0.59*RE = 0.59*sqrt(2*GM/RE)

initial energy on the earth Ei = (1/2)*m*v^2 - G*M*m/RE

at the maximum point final energy Ef = -G*M*m/r

from energy conservation

Ef = Ei

-G*M*m/r = -G*M*m/RE + (1/2)*m*0.59^2*(2*G*M/RE)

-1/r = -1/RE + 0.59^2/RE

-1/r = -0.6519/RE

r = 1.53xRE <<<<----------answer

===================

b)

if projectile speed v = 0.33*RE = 0.33*sqrt(2*GM/RE)

initial energy on the earth Ei = (1/2)*m*v^2 - G*M*m/RE

at the maximum point final energy Ef = -G*M*m/r

from energy conservation

Ef = Ei

-G*M*m/r = -G*M*m/RE + (1/2)*m*0.59^2*(2*G*M/RE)

-1/r = -1/RE + 0.33^2/RE

-1/r = -0.8911/RE

r = 1.122xRE <<<<----------answer

===================

c)

least energy E = (1/2)*m*vE^2 - G*M*m/RE

E = (1/2)*m*2*G*M/RE - G*M*m/RE

E = -G*M*m/(2*RE)

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