A satellite with mass $\mathbf{848}\mathbf{}\mathbf{Kg}$is in a circular orbit with an orbital speed of$\mathbf{9640}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$around the earth. What is the new orbital speed after friction from the earth’s upper atmosphere has done$\mathbf{-}\mathbf{7}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{\text{\hspace{0.17em}J}}$of work on the satellite? Does the speed increase or decrease?

The given data can be listed below as,

• The mass of satellite is,$m=848\text{\hspace{0.17em}kg}$.
• The speed of the satellite is,$v_1=9640\text{\hspace{0.17em}m}/\text{s}$.
• The work is,$W_0=-7.50\times {10}^9\text{\hspace{0.17em}J}$.

The principle is that energy is neither formed nor demolished. It just moves from one place to another energy type.

The conservation of the energy is expressed as,

$K_1+U_1+W_0=K_2+U_2$ ...(i)

Here$W_o$is the engine do work on the rocket,$U_2$is the gravitational potential energy for an orbital second, $U_1$is the gravitational potential energy for an orbital first orbital,$K_2$is the kinetic energy for an orbital second,$K_1$is the kinetic energy for orbital first orbital.

The relation of gravitational potential energy is expressed as,

$U=-\frac{Gm{M}_E}{r}$ ...(ii)

Here$U$is the potential energy,$G$is the gravitational constant,$M_E$is the mass of the earth,$m$is the mass of the satellite, and$r$is the radius of the orbit.

The satellite move in a circular orbit, and then the speed of the satellite is expressed as,

$v=\sqrt{\frac{G{M}_E}{r}}$

Here$v$is the speed of the satellite.

The relation of kinetic energy is expressed as,

$K=\frac{1}{2}m{v}^2$ ...(iii)

Here K is the kinetic energy and m is the mass of the satellite.

Substitute the value$v$in equation (iii)

$\begin{array}{c}K=\frac{1}{2}m{\left(\sqrt{\frac{G{M}_E}{r}}\right)}^2\\ =\frac{Gm{M}_E}{2r}\end{array}$

The value of$-\frac{Gm{M}_m}{r}$is equal to the gravitational potential energy and then the kinetic energy is expressed as,

$\begin{array}{l}K=-\frac{1}{2}U\\ U=-2K\end{array}$

Then the gravitational potential energy for the first orbit is expressed as,

$U_1=-2K_1$

Here$U_1$is the potential energy and$K_1$is the kinetic energy of the first orbit.

The gravitational potential energy for the second orbit is expressed as,

$U_2=-2K_2$

Hererole="math" localid="1655791042622" $U_2$is the potential energy androle="math" localid="1655791051428" $K_2$is the kinetic energy of the second orbit.

The kinetic energy for the first orbit is expressed as,

$K_1=\frac{1}{2}m{v}_1^2$

Here$v_1$is the potential energy and$K_1$is the kinetic energy of the first orbit.

The kinetic energy for the second orbit is expressed as,

$K_2=\frac{1}{2}m{v}_2^2$

Here$v_2$ is the potential energy and$K_2$ is the kinetic energy of the second orbit.

Substitute the value of $K_1$,$K_2$ ,$U_1$ , and $U_2$, in the equation (i).

$\begin{array}{c}{K}_1-2K_1+W_0=K_2-2K_2\\ -K_1+W_0=-K_2\\ -\frac{1}{2}m{v}_1^2+W_0=-\frac{1}{2}m{v}_2^2\\ v_2^2=v_1^2-\frac{2W_0}{m}\\ v_2=\sqrt{v_1^2-\frac{2W_0}{m}}\end{array}$

Substitute$9640\text{\hspace{0.17em}m}/\text{s}$ for$V_1$ , $848\text{\hspace{0.17em}kg}$for$m$ , and $-7.50\times {10}^9\text{\hspace{0.17em}J}$for$W_O$ in the above equation.

$\begin{array}{c}{v}_2=\sqrt{{(9640\text{\hspace{0.17em}m}/\text{s})}^2-\frac{2\times (-7.50\times {10}^9\text{\hspace{0.17em}J})}{848\text{\hspace{0.17em}kg}}}\\ =10517\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed is increased and it value is, $10517\text{\hspace{0.17em}m}/\text{s}$.