91Ó°ÊÓ

Mass of satellite A (M_a) and Mass of satellite B${\mathrm{M}}_{\mathrm{a}}={\mathrm{M}}_{\mathrm{b}}=\text{m}=14.6\text{ }\mathrm{kg}$.

Height of satellite A${\text{h}}_{\text{a}}=6370\text{ }\mathrm{km}\left(\frac{{10}^3\text{ }\mathrm{m}}{1\text{ }\mathrm{km}}\right)=6370×{10}^3\mathrm{m}$

Height of satellite B${\text{h}}_{\text{b}}=19110\text{ }\mathrm{km}\left(\frac{{10}^3\text{ }\mathrm{m}}{1\text{ }\mathrm{km}}\right)=19110×{10}^3\mathrm{m}$

The radius of the Earth ${\text{R}}_{\text{E}}=6370\text{ }\mathrm{km}\left(\frac{{10}^3\text{ }\mathrm{m}}{1\text{ }\mathrm{km}}\right)=6370×{10}^3\mathrm{m}$

By using potential energy and kinetic energy formulae, we can find out

$\left(\frac{{\text{U}}_{\text{b}}}{{\text{U}}_{\text{a}}}\right)$And$\left(\frac{\text{Kb}}{{\text{K}}_{\text{a}}}\right)$

By adding potential energy and kinetic energy, we can find the total energy.

Formula:

$\mathrm{U}=\frac{−\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$

$\mathrm{K}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}$

$\mathrm{E}=\mathrm{U}+\mathrm{K}$

$\mathrm{E}=−\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}$

$\left(\frac{{\text{U}}_{\text{b}}}{{\text{U}}_{\text{a}}}\right)$

${\text{U}}_{\text{b}}\text{=}\frac{\text{-GMm}}{{\text{R}}_{\text{E}}{\text{+h}}_{\text{b}}}$ (1)

${\text{U}}_{\text{a}}\text{=}\frac{\text{-GMm}}{{\text{R}}_{\text{E}}{\text{+h}}_{\text{a}}}$ (2)

Dividing equations 1 and 2

$\frac{{\text{U}}_{\text{b}}}{{\text{U}}_{\text{a}}}\text{=}\frac{\text{-GMm}}{{\text{R}}_{\text{E}}\text{+hb}}\text{×}\frac{{\text{R}}_{\text{E}}\text{+ha}}{\text{-GMm}}$

$\begin{array}{c}\frac{{\text{U}}_{\text{b}}}{{\text{U}}_{\text{a}}}\text{=}\frac{{\text{R}}_{\text{E}}\text{+ha}}{{\text{R}}_{\text{E}}\text{+hb}}\\ \text{=}\frac{\text{6370 k³¾+6370 k³¾}}{\text{6370 k³¾+19110 k³¾}}\end{array}$

$\frac{{\text{U}}_{\text{b}}}{{\text{U}}_{\text{a}}}=\frac{1}{2}$

${\text{U}}_{\text{a}}=2{\text{U}}_{\text{b}}$ (3)

Calculation for$\left(\frac{{\text{K}}_{\text{b}}}{{\text{K}}_{\text{a}}}\right)$

${\text{K}}_{\text{b}}\text{=}\frac{\text{GMm}}{\text{2}\left({\text{R}}_{\text{E}}{\text{+h}}_{\text{b}}\right)}$ (4)

${\text{K}}_{\text{a}}\text{=}\frac{\text{GMm}}{\text{2}\left({\text{R}}_{\text{E}}{\text{+h}}_{\text{a}}\right)}$ (5)

Dividing equation (4) by (5)

$\frac{{\text{K}}_{\text{b}}}{{\text{K}}_{\text{a}}}\text{=}\frac{{\text{R}}_{\text{E}}{\text{+h}}_{\text{a}}}{{\text{R}}_{\text{E}}{\text{+h}}_{\text{b}}}$

$\frac{{\text{K}}_{\text{b}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}}{\text{2}}$

${\text{K}}_{\text{a}}{\text{=2K}}_{\text{b}}$

Comparing the total energy of satellite A and satellite B

The satellite with a smaller value of the radius of orbit would have larger energy. Also, the value of E is negative. So, the satellite with the smallest value of magnitude would have the largest energy. Therefore, satellite B has the largest energy.

Calculation for${\mathrm{E}}_{\mathrm{a}}$and${\mathrm{E}}_{\mathrm{b}}$

$\begin{array}{c}{\text{E}}_{\text{a}}=−\frac{\text{GMm}}{\text{2}\left({\text{R}}_{\text{E}}{\text{+h}}_{\text{a}}\right)}\\ =−\frac{6.67×{10}^{−11}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2×5.98×{10}^{24}\text{ }\mathrm{kg}×14.6\text{ }\mathrm{kg}}{2(6370+6370){10}^3\text{ }\mathrm{m}}\\ =−2.2×{10}^8\mathrm{J}\end{array}$

$\begin{array}{c}{\text{E}}_{\text{b}}=\frac{{\text{E}}_{\text{a}}}{\text{2}}\\ =\frac{−2.2×{10}^8\mathrm{J}}{2}\\ =−1.1×{10}^8\mathrm{J}\end{array}$

$\begin{array}{c}\left|{\text{E}}_{\text{a}}\right|\text{-}\left|{\text{E}}_{\text{b}}\right|=2.2×{10}^8\mathrm{J}−1.1×{10}^8\mathrm{J}\\ =1.1×{10}^8\mathrm{J}\end{array}$

${\mathrm{E}}_{\mathrm{a}}$ Is greater than ${\mathrm{E}}_{\mathrm{b}}$ by $1.1×{10}^8\mathrm{J}$