Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of
R/2 where
R is the radius of the Earth. If
F(r) is the gravitational force of the Earth on a point mass at a distance
r, what is:
\frac{F(R)}{F(2R)}?
- 32
- 8
- 4
- 2
- 1
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Using the same logic, (1/2)^2 is 1/4. Use basic math, cross out the k's and R's, and tada: Just 1/4.
Yup that's correct.
Careful, the gravitational force is not inverse square inside the earth, if it were you would have a nearly infinite force at the center of the earth and the planet would implode.
Ooops. I guess I wasn't thinking very straight. The correct answer is 2 isn't it?
By Gauss's Law for Gravity,
g\cdot A_{enclosed} = 4\pi G\rho V_{enclosed}. (Some will note that I have reduced it to a form that is useful for this question.) At a radius of r, the (surface) area enclosed is: A_{enclosed} = 4\pi r^2. Likewise, the volume enclosed is: V_{enclosed} = \frac{4}{3}\pi r^3. Finally, the density of the Earth, \rho, is: \rho = \frac{M}{\frac{4}{3}\pi R^3} where M and R are the radius and mass of the Earth respectively.
Putting it all in, we have: g\cdot (4\pi r^2) = 4\pi G\frac{M}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3. After some simplification, we have g = \frac{GM}{2R^2} at r = \frac{R}{2}. Recall that at r=R, g=\frac{GM}{R^2}. Dividing the two expressions for g above gives 2 as the proper answer.
for the record, the answer for \frac{F(R)}{F(R/2)} is *not* 4.
The second answer's going to be 1/4, isn't it?
Using the same logic, (1/2)^2 is 1/4. Use basic math, cross out the k's and R's, and tada: Just 1/4.
Yup that's correct.
Careful, the gravitational force is not inverse square inside the earth, if it were you would have a nearly infinite force at the center of the earth and the planet would implode.
Ooops. I guess I wasn't thinking very straight. The correct answer is 2 isn't it?
Let me take this time to explain it properly and redeem myself a little.
By Gauss's Law for Gravity,
g\cdot A_{enclosed} = 4\pi G\rho V_{enclosed}. (Some will note that I have reduced it to a form that is useful for this question.) At a radius of r, the (surface) area enclosed is: A_{enclosed} = 4\pi r^2. Likewise, the volume enclosed is: V_{enclosed} = \frac{4}{3}\pi r^3. Finally, the density of the Earth, \rho, is: \rho = \frac{M}{\frac{4}{3}\pi R^3} where M and R are the radius and mass of the Earth respectively.
Putting it all in, we have: g\cdot (4\pi r^2) = 4\pi G\frac{M}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3. After some simplification, we have g = \frac{GM}{2R^2} at r = \frac{R}{2}. Recall that at r=R, g=\frac{GM}{R^2}. Dividing the two expressions for g above gives 2 as the proper answer.
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