EK Reasoning Skills #3 (pg.20)

[aspiringdocO]

The force of gravity on any object due to the earth is given by the equation F =(GmM/r^2), where (blah, blah, blah) and r is the distance of between the center of the mass of the earth and the center of the mass of the object. If a rocket weighs 3.6x 10^6 N at the surface of the earth, what is the force on the rocket due to the gravity when the rocket has reached an altitude of 1.2x 10^4km? (radius of earth= 6370 km)

A. 1.2x10^5N
B.4.3x10^5N
C. 4.8x10^6N
D.9.6x10^6N

EK's explanation: "If you are good with scientific notation, it is easy to see that r is tripled. r is the distance from the center of the Earth to the earth's surface. The Satellite is 2 Earth radii from the surface of the Earth so it is three earth radii from the center of the earth. Since the square of r is inversely proportional to F, F must be divided by 9."

Can somebody help me see HOW r is being tripled? I'm having a brain fart or something...

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[Labrat07]

r = center of one planet to the center of other planet. Make sure you draw this out. So if you have a planet with 2r from center to one surface. You still need to add the radius of that surface planet. In this case = r. r+2r = 3r

 

[aspiringdocO]

But can you explain how I could know r is tripled using scientific notation (the numbers given)?

 

[Labrat07]

ok, so r of earth = 6370. Distance from surface earth to center other planet = 12k. 12k/6370 = close to 2 (smaller). so we have r +2r = 3 r
(3r) ^2 = 9. The new force will be a ratio of 1/9 of old one

3.6 x 10^6 / 9 = 4 x10^ 5 ish.