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Question

The period TT of a satellite circling Earth is given by T2=kR3(1+dR)3,T^{2}=k R^{3}\left(1+\frac{d}{R}\right)^{3}, where RR is the radius of Earth, dd is the distance of the satellite above Earth, and kk is a constant. Solve for R,R, using fractional exponents in the result.

Solution

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The period TT is given by the following equation:

T2=kR3(1+dR)3T^2=kR^3\left( 1+\dfrac{d}{R}\right)^3

Take the cube root of both sides of the equation as follows:

T23=kR3(1+dR)33\sqrt[3]{T^2}=\sqrt[3]{kR^3\left( 1+\dfrac{d}{R}\right)^3}

T23=k3R33(1+dR)3\sqrt[3]{T^2}=\sqrt[3]{k} \sqrt[3]{R^3} \sqrt[3]{\left( 1+\dfrac{d}{R}\right)}

Apply this exponent rule ann=a\sqrt[n]{a^n}=a as follows:

T23=k3R(1+dR)\sqrt[3]{T^2}=\sqrt[3]{k} R \left( 1+\dfrac{d}{R}\right)

Remove the parentheses by multiply k3\sqrt[3]{k} by the inside terms as follows:

T23=(k3R+dk3RR)\sqrt[3]{T^2}= \left( \sqrt[3]{k} R+\dfrac{d\sqrt[3]{k} R}{R}\right)

Divide the common factor RR as follows:

T23=k3R+dk3\sqrt[3]{T^2}= \sqrt[3]{k} R+d \sqrt[3]{k}

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