TBR gravitational ratio question

[ohitsthatguy]

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Question number 11 from the Force, Circular Motion, Gravity Test in TBR Physics 1 Book
If the space station is located a distance of twice the radius of the Earth away from the surface of the Earth, the acceleration due to gravitational attraction from the Earth on the space station is:
A. one-ninth of that on Earth.
B. one-quarter of that on Earth.
C. one-half of that on Earth.
D. equal to that on Earth.
ANSWER: A

I don't understand how they reached the point where they said that the radius is 3RE? I plugged it in as twice the distance because that is what the question stated.

Explanation from book given:
To solve for the acceleration (call it g') exerted on the space station from the Earth, we use:
g'= GME/r^2
where r is the center-to-center distance between the Earth and the space station. This is a spin-off of the gravitational force equation, r for this problem is 3RE, where RE is the radius of the Earth. (Note that they stated the space station was 2RE away from the surface of the Earth. Remember to add in the radius of the Earth to get the center-to-center distance.) We do not know the mass of the Earth, nor do we know the radius of the Earth, so the best way to solve this problem is to recognize that g, the acceleration due to gravity at the Earth's surface, can be written as:
g= GME/(R^2E)
We then take the ratio of g' to g: GME/(3RE)2 : GME/(RE)^2= 1/9

 

[siffster]

" ...twice the radius of the Earth away from the surface of the Earth..."

RE=radius of the earth
so the satellite is 2RE away from the SURFACE of the earth. remember the surface of the earth is 1RE away from the center of earth. When dealing with F=GMm/r^2 problems, the value for r in the denominator is always measured from the center of the earth. so if the satellite would be at a distance H away from the surface of the earth, the total distance separating the satellite and the center of the earth would be RE+H. You would have to plug in F=GMm/(RE+H)^2 to find the force that they exert on each other

so for this problem, the satellite is a distance 2RE away from the surface of the Earth, but 3RE away from the center.

hope that helped

 

[ohitsthatguy]

Oh wow.... That makes sense, I totally disregarded the surface part of it. Thank you so much!

 

[brood910]

Dont bother with finding the ratio.
Just multiply r by 3 times in the equation. What happens next?

 

[ohitsthatguy]

Then the force would be 1/9th because you square it? Yea you are right, the ratio seems like it is just over 1 really.