Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed $\mathbf{2}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$when at a distance of$\mathbf{2}\mathbf{.5}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{\text{\hspace{0.17em}m}}$from the center of the sun, what is its speed when at a distance of$\mathbf{5}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{\text{\hspace{0.17em}m}}$?

The given data can be listed below as,

• The speed of the comet is,$v_1=2.0\times {10}^4\text{\hspace{0.17em}m}/\text{s}$.
• The distance of the comet from the center of the sun is, $r_1=2.5\times {10}^{11}\text{\hspace{0.17em}m}$.
• The distance of the comet from the center of the sun is, $r_2=5.0\times {10}^{11}\text{\hspace{0.17em}m}$.
• The mass of the sun is,$M_s=1.9891\times {10}^{30}\text{\hspace{0.17em}kg}$

Gravity refers to the force of attraction between two objects. Gravity refers to the force of attraction between the Earth and other objects and its depends upon the amount of matter present in the objects.

Apply the conservation of the energy between sun and comet. It is expressed as,

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

Here $K_1$, $K_2$and$U_1$,$U_2$are the kinetic and gravitational potential energy, androle="math" localid="1655730398445" $W_o$is the other work.

The sun's gravity is the only force acting on a comet, so $W_o=0$.

Substitute the value of $W_o$in the equation (i).

$K_1+U_1=K_2+U_2$ ...(ii)

The kinetic energy and gravitational potential energy at a distance$2.5\times {10}^{11}\text{\hspace{0.17em}m}$.it is expressed as,

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

Here $m$is the mass of comet. And,

$U_1=-G\frac{M_{s}m}{r_1}$

Here$G$ is the gravitational constant,$M_s$is the mass of sun and$m$is the mass of comet.

The kinetic energy and gravitational potential energy at a distance $5.0\times {10}^{10}\text{\hspace{0.17em}m}$.it is expressed as,

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

And

$U_2=-G\frac{M_{s}m}{r_2}$

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

$\begin{array}{c}\frac{1}{2}m{v}_1^2-G\frac{M_{s}m}{r_1}=\frac{1}{2}m{v}_2^2-G\frac{M_{s}m}{r_2}\\ v_2^2=v_1^2+2G{M}_s\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ =v_1^2+2G{M}_s\left(\frac{r_1-r_2}{r_1{r}_2}\right)\end{array}$

Substitute$6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for$G$ , $1.9891\times {10}^{30}\text{\hspace{0.17em}kg}$for$M_s$ ,$2.5\times {10}^{11}\text{\hspace{0.17em}m}$ for$r_1$ , $5.0\times {10}^{10}\text{\hspace{0.17em}m}$ for $r_2$and $2.0\times {10}^4\text{\hspace{0.17em}m}/\text{s}$for$v_1$ in the above equation.

$\begin{array}{c}{v}_2^2={(2.0\times {10}^4\text{\hspace{0.17em}m}/\text{s})}^2+2\times 6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 1.9891\times {10}^{30}\text{\hspace{0.17em}kg}\times \left(\frac{2.5\times {10}^{11}\text{\hspace{0.17em}m}-5.0\times {10}^{10}\text{\hspace{0.17em}m}}{2.5\times {10}^{11}\text{\hspace{0.17em}m}\times 5.0\times {10}^{10}\text{\hspace{0.17em}m}}\right)\\ v_2=6.8\times {10}^4\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed at a distance$5.0\times {10}^{10}\text{\hspace{0.17em}m}$ is, $6.8\times {10}^4\text{\hspace{0.17em}m}/\text{s}$.