GREPhysics.NET: GR8677 #75

GR8677 #75

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }Kepler's Third Law

One recalls (or should memorize now) the famous slogan: The square of the period is proportional to the cube of the distance. In equation form,

$T^2 = k d^3<br />$

One does not need to know the proportionality factor to solve this problem. In fact, the problem gives $T=80$ minutes at $d=R_e$ . Thus, $80^2=k R_e^3 \Rightarrow k = 80^2minutes/R_e^3$ .

There are $24 \times 60$ minutes in 24 hours. Plug that into the equation (and the relation for $k$ as determined above) to get,

$(24 \times 60)^2 = 80^2/R_e^3 d^3 \Rightarrow d=(24 \times 6/8)^{2/3} R_e<br />$

No calculators allowed, so $(18^2)^{1/3}$ is about $400^{1/3}$ , which is closest to $7$ , as in choice (B).

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Comments

herrphysik
2006-10-30 00:32:29

No need to solve for a proportionality factor, just use the form $(T_1/T_2)^2=(s_1/s_2)^3$ . Same thing. Also, the exact result I get is $4.2R_e$ , so I suppose it would actually be $7R_e$ and not $3R_e$ due to the gravitational force being less at the higher orbit?

tin2019
2007-10-29 10:52:42

The exact result is 6.868

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