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


  1. 32
  2. 8
  3. 4
  4. 2
  5. 1

Mechanics}Gravitational Law

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

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


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\nNEC
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
NEC
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.
Typo Alert!
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)\piR^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
NEC

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Don't you have to take into account that the mass is not concentrated at the center? where \rho = M / (4/3)\piR^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?

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