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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.

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter13: Gravitation
Section: Chapter Questions
Problem 27P: The acceleration due to gravity on the surface of a planet is three times as large as it is on the...
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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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