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Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force F acting on an object situated on this line segment is F=Kx2+0.012K(239x)2F=\frac{-K}{x^{2}}+\frac{0.012 K}{(239-x)^{2}} where K>0 is a constant and x is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the "dead spot' where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.)


Answered 4 months ago
Answered 4 months ago
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Let xx be the distance from the center of the earth to the dead spot (in thousands of miles).

Setting F=0F=0, we have

0=Kx2+0.012K(239x)20=-\displaystyle \frac{K}{x^{2}}+\frac{0.012K}{(239-x)^{2}}

Kx2=0.012K(239x)2\displaystyle \frac{K}{x^{2}}=\frac{0.012K}{(239-x)^{2}}



57121478x+x2=0.012x257121 -478x+x^{2}=0.012x^{2}\\\\$0.988x^{2}-478x+57121=0.\

Usingthequadraticformula,Using the quadratic formula, \\\\x=\displaystyle (478)±(478)24(0.988)(57121)2(0.988)\frac{-(-478)\pm\sqrt{(-478)^{2}-4(0.988)(57121)}}{2(0.988)}


=\displaystyle 478±2741.8081.976\frac{478\pm\sqrt{2741.808}}{1.976}\approx478±52.3621.976\frac{478\pm 52.362}{1.976}\approx 241.903\pm 26.499.

x_{1}\approx 241.903+26.499\approx 268,         x_{2}\approx 241.903-26.499\approx 215.\

Since268(000miles)isgreaterthanthedistancefromtheearthtothemoon,werejectit.So,Since 268(000 miles) is greater than the distance from the earth to the moon, we reject it.\\\\ So,\\x\approx 215,000$ miles.

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