Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
5
Circular Motion; Gravitation
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5-1 to 5-3: Uniform Circular Motion; Highway Curves
5-4: Nonuniform Circular Motion
5-6 and 5-7: Law of Universal Gravitation
5-8: Satellites; Weightlessness
5-9: Kepler's Laws

Problem 31
A
gp=4.4m/s2g_p = 4.4m/s^2
Giancoli 6th Edition, Chapter 5, Problem 31 solution video poster
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VIDEO TRANSCRIPT

Here’s the formula we’re going to use to calculate the acceleration due to gravity on the planet ;’g’ equals ‘G’ times mass of the planet ‘mp’ divided by the radius ‘rp’ of the planet squared. The mass of the planet we’re told is the same as he mass of the earth ‘mE’ and its radius is going to be one point five times the radius of the earth ‘rE’. So acceleration due to gravity there ‘gp’ equals ‘G’ times ‘mE’ divided by one point five times ‘rE’ all squared which is one over one point five squared times ‘G’ times ‘mE’ over ‘rE’ squared and I did it this way because we could’ve substituted for the mass and radius of the earth both of which we can find in the front cover of the textbook but when we write it as I’ve shown we can see ‘G’ times ‘mE’ over ‘rE’ squared is going to be ‘g’ which we’re used to, nine point nine eight that is ‘g’ on the earth making the calculation a little bit simpler we don’t need such big numbers and we’re going to say that this is one over one point five squared times acceleration due to gravity on the earth. So that’s nine point eight divided by two point two five which is four point four meters per second squared. In the 5th Edition the radius of the planet is two point five times the radius of the earth so the acceleration due to gravity to this planet is one point six meters per second squared.

COMMENTS
By jamesmontyorourke on Wed, 12/9/2020 - 12:09 PM

Bless you mr Giancoli

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