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 the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.712 of the escape speed from Earth and (b) its initial kinetic energy is 0.712 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Explanation / Answer

a) Escape velocity , Ve = sqrt(2GM/RE)

V = 0.712 Ve

Using Conservation of Energy

PE1 + KE1 = PE2 + KE2

- GMm / RE  + 0.5 m (V)2  = - GMm/r + 0

Put V = 0.712 *sqrt(2GM/RE)

- GMm/RE + 0.5*0.712^2*2 GMm/RE = - GMm/r

-1/RE + 0.5069/RE = - 1/r

r =RE /(1-0.5069) = 2.028 RE

a) multiple of radius = 2.028

b) Given that 0.5 mV^2 = 0.712*0.5*2GMm/RE

Using Conservation of Energy

PE1 + KE1 = PE2 + KE2

- GMm / RE  + 0.5 m (V)2  = - GMm/r + 0

- GMm/RE + 0.5*0.712*2 GMm/RE = - GMm/r

-1/RE + 0.712/RE = - 1/r

r = RE / (1-0.712) = 3.472

multiple of radius = 3.472

c)

When the projectile reaches infinity, it stops and thus has no kinetic energy and potential
. Its total mechanical energy at infinity is zero

Hence answer is zero

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a) 2.028

b) 3.472

c) 0

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