240 CHAPTER 13 PRACTICE GRAVITY Guided Problem 13.6 More rocket launches Three r

Question

240 CHAPTER 13 PRACTICE GRAVITY Guided Problem 13.6 More rocket launches Three rockets are launched from Earth as shown in Figure WG13.5. One rocket is launched vertically (a), one horizontally (b), and one at 45° with respect to the ground (c). All three are launched with a speed of the escape speed from Earth's surface, tesc-1.12 × 104 m/s (from Principles Checkpoint 13.22).Calculate the maximum distance from Earth each rocket achieves. Ignore Earth's rotation, and express your answer in terms of Earth's mass and radius. Figure WG13.5 45°

Explanation / Answer

given v = 3vesc/4
ves = 1.12*10^4 m/s
hence
v = 8.4*10^3 m/s
let mass be m

now, case a. initial gravitational potential energy = -GMm/d
here M = mas sof earth = 5.972*10^24 kg
d = radius of earth = 6371,000 m
hence PEi = -GMm/d = -62522743.6823m J
let the space craft reach an altitude of h
then
from conservation fo energy
-62522743.6823m + 0.5*m*v^2 = -GMm/(h + d)
h = 8250596.29 m

case b. here we can assume the angular momentum of the satellite will remain constant
its speed at altitide h is u
then
mv*d = m*u*(d + h)
u = v*d/(d + h)
hence
from conservaiton of energy
-62522743.6823m + 0.5*m*v^2 = -GMm/(h + d) + 0.5m*v^2*d^2/(d + h)^2
solving this quadratic equation will give us the answer

similiary for case c. from conservation of angular moemntum
m*v*dcos(45) = m*u*(d + h)
u = v*d*cos(45)/(d + h)
hence
from conservaiton of energy
-62522743.6823m + 0.5*m*v^2 = -GMm/(h + d) + 0.5m*v^2*d^2*cos^2(45)/(d + h)^2
solving this quadratic equation will give us the answer

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