GREPhysics.NET: GR9277 #4

GR9277 #4

Problem

GREPhysics.NET Official Solution

\prob{4}
The magnitude of the Earth's gravitational force on a point mass is F(r), where r is the distance from the Earth's center to the point mass. Assume the Earth is a homogenous sphere of radius R.

What is $\frac{F(R)}{F(2R)}$

Mechanics $\Rightarrow$ }Gravitational Law

Recall the famous inverse square law determined almost half a millennium ago,

$F=\frac{k}{r^2}, <br />$

where $k=GMm$ .

The ratio of two inverse-square forces ( $r>R$ , where $R$ is the radius of the planet or huge heavy object) would be

$\frac{F(r_1)}{F(r_2)}=\frac{4r_2^2}{r_1^2}.<br />$

Thus, $\frac{F(R)}{F(2R)}=\frac{4R^2}{R^2}=4$ , which is choice (C).

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Comments

bonghan_lee27
2016-05-22 17:07:38

Easy question. 97% obtained it right.\r\n

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Manoj
2011-06-20 23:15:12

g/g'= r'*r'/r*r
Substituting the respective values the answer will be 4.

dwight5
2018-05-14 04:08:42

Agree!\r\ngmail sign up

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solarclathrate
2009-06-22 08:38:19

Should not the LHS of the ratio of forces equation in the solution read $\frac{F(r_1)}{F(2 r_2)}$ ?

UTBphysics
2009-10-11 01:56:19

not exactly since R=r1 and r2=2R=2r1 the LHS is alright. The r2 on the RHS should be an r1, however.

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emailzac
2007-11-02 16:11:10

Don't you have to take into account that the mass is not concentrated at the center? where $\rho$ = M / (4/3) $\pi$ R^3 you get a mass of M/8 then take into account the radial contribution of 4 to get a final answer of 1/2. Where'd i go wrong?

emailzac
2007-11-02 16:15:28

Sorry, thought this was problem 5, please ignore this comment :D

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Easy question. 97% obtained it right.\r\n

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