The answer is already shown, I am just confused about the steps. On part a, how does the 3.1447 go to 4.94Re? and then h=3.94Re and then 25097800m? On part b, where does the 2.5 come from? please solve in simpler steps, thank you. 4) [Ignoring air resistance], a giant slingshot gives a pebble a velocity of 10,000 m/sec at the surface of the Earth.(GM=4x1014 Nm²/kg, Re=6.37x10°m, v-esc=11200 m/s). [No, you don't have the mass of the pebble, and youdon't need it.]a) Calculate the maximum height above the surface of the Earth that the pebble will rise (in meters).b) When the pebble is 1.5 Earth radii above the surface of the Earth, calculate how fast it is going (in m/sec).2GM2(4x10*)v -v 11200² – 10000²2GMa) ジー。= 3.1447x10' m= 4.94R, h=3.94R, = 25,097,800mr =レesc2GM2GM1らー=2.5RĘescescb)2.52.5rv = 10000² +(-0.6)11200² / v, = 4973m / s

Answered: Image /qna-images/answer/497266ea-b2f4-4d00-9fd0-06866be6fb5c.jpg

The answer is already shown, I am just confused about the steps. On part a, how does the 3.1447 go to 4.94Re? and then h=3.94Re and then 25097800m? On part b, where does the 2.5 come from? please solve in simpler steps, thank you.

The answer is already shown, I am just confused about the steps. On part a, how does the 3.1447 go to 4.94Re? and then h=3.94Re and then 25097800m? On part b, where does the 2.5 come from? please solve in simpler steps, thank you.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter13: Universal Gravitation

Section: Chapter Questions

Problem 13.73AP

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Question

4) [Ignoring air resistance], a giant slingshot gives a pebble a velocity of 10,000 m/sec at the surface of the Earth.
(GM=4x1014 Nm²/kg, Re=6.37x10°m, v-esc=11200 m/s). [No, you don't have the mass of the pebble, and you
don't need it.]
a) Calculate the maximum height above the surface of the Earth that the pebble will rise (in meters).
b) When the pebble is 1.5 Earth radii above the surface of the Earth, calculate how fast it is going (in m/sec).
2GM
2(4x10*)
v -v 11200² – 10000²
2GM
a) ジー。
= 3.1447x10' m= 4.94R, h=3.94R, = 25,097,800m
r =
レ
esc
2GM
2GM
1
らー=
2.5RĘ
esc
esc
b)
2.5
2.5
r
v = 10000² +(-0.6)11200² / v, = 4973m / s

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