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Ber [7]
3 years ago
7

The gravitational force, F, on a rocket at a distance, from the center of the earth is given by F = k / (r^2), where k = 10 ^3 n

ewton*km^2. when the rocket is 10^4 km from the center of the earth, its moving away at .2 km/sec. how fast is the gravitational force changing at that moment? therefore what is the rate of change of the gravitational force in newton/ second?
Mathematics
1 answer:
makvit [3.9K]3 years ago
3 0
F = k / (r^2)
dF/dt = -2k / (r^3) dr/dt

When r = 10^4 km and dr/dt = 0.2 km/sec
dF/dt = -2(10)^3 (0.2) / ((10^4)^3) = -400 / 10^7 = -0.00004 N/s

The gravitational force is changing at the rate of -0.00004 N/s
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\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\
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\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
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4 0
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ELEN [110]
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now we sub
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a(8) = 12582912 <==
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