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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.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 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
Transcribed Image Text: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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