91Ó°ÊÓ

The given data can be listed below,

• The radius of the planet is, $R=9.00×{10}^{6}m$.
• The mass of first satellite one is, $m_1=68.0\mathrm{kg}$.
• The radius of first satellite one is, $r_1=7.0×{10}^{7}m$.
• The orbital speed of first satellite is, $v_1=4800m/s$.
• The mass of second satellite is, $m_2=84kg$.
• The radius of second satellite is, $r_2=3.00×{10}^{7}m$

The speed of a satellite in a circular path keeps the distance between it and the Earth's center consistent, signifying that the satellite is traveling in a uniform circular motion.

Newton’s second law acting on the satellites is given by,

$F=\frac{m{v}^2}{r}$

Here,m is the mass of satellite, v is the orbital speed of satellite and r is the radius of satellite.

The gravitational law acting on satellites is given by,

$F_g=\frac{GMm}{r^2}$

Here,Mis the mass of the planet whose value is $5.97×{10}^{24}\mathrm{kg}$, G is the constant of gravitation whose value is $G=6.673×{10}^{-11}N·m/k{g}^2$,m is the mass of the satellite, r is the radius of satellite and v is the orbital speed of the satellite.

So from law of conservation of energy,

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$

Solve the above equation for constant term.

$GM=r{v}^2$

From the above equation.

$\begin{array}{r}{r}_1{V}_1^2=r_2{v}_2^2\\ V_2=V_1\sqrt{\frac{r_1}{r_2}}\end{array}$

Here,$v_1$ is the orbital speed of first satellite,$r_1$ is the radius of first satellite,$r_2$ is the radius of second satellite.

Substitute all the values in the above expression.

$\begin{array}{r}{v}_2=(4800\mathrm{m}/\mathrm{s})\sqrt{\frac{7.00×{10}^7\mathrm{m}}{3.00×{10}^7\mathrm{m}}}\\ =7330\mathrm{m}/\mathrm{s}\end{array}$

Thus, the orbital velocity of the second satellite is 7330 m/s.