## Related questions with answers

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

Solution

VerifiedThe period $T$ is given by the following equation:

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

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

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

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

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

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

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

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

Divide the common factor $R$ as follows:

$\sqrt[3]{T^2}= \sqrt[3]{k} R+d \sqrt[3]{k}$

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