91Ó°ÊÓ

The given data can be listed below,

The gravitational force exerted by the star on the planet is,$F\_\{\backslash mathrm\{SP\left\}\right\}=4\backslash times10^\left\{25\right\}\backslash mathrm\{~N\}.$

From the above equation. The gravitational force on the star by the planet is given by,

$\backslash begin\left\{aligned\right\}F\_\{PS\}\&=-F\_\{SP\}\backslash \backslash \&=-4\backslash times10^\left\{25\right\}\backslash mathrm\{~N\}\backslash end\left\{aligned\right\}$

Thus, the gravitational force on the star by the planet is $-4×{10}^{25}\mathrm{N}$.

The conservation of momentum is due to the fact that when your frame of reference is switched, the rules of physics stay intact, as expressed by Newton's third law of motion.

(a)

By using Newton's third law, force on two points can be equated as,

$F_{XY}=-F_{YX}$

Here, $F_{XY}$is the force exerted by $X$on $Y$and $-F_{YX}$is the force exerted by $Y$on $X$in the opposite direction.

(b)

The gravitational force on the planet by the star is given by,

$F_{SP}=\frac{G{m}_s{m}_P}{r^2}$

Here, $G$is a universal gravitational constant whose value is $6.673×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2$$m_s$is the mass of the star, $m_P$is mass of the planet, and $r$is the distance between the centers of the planets.

When the mass of the planet increases by two times, the magnitude of the force is given by,

$F_{SP}^{\text{'}}=\frac{G\left(2m_s\right)m_P}{r^2}$

Substitute all the values in the above,

$\backslash begin\left\{aligned\right\}F\_\{SP\}^\{\backslash prime\}\&=2F\_\{SP\}\backslash \backslash \&=2\backslash left(4\backslash times10^\{25\}\backslash mathrm\{~N\}\backslash right)\backslash \backslash \&=8\backslash times10^\left\{25\right\}\backslash mathrm\{~N\}\backslash end\left\{aligned\right\}$

Thus, the magnitude of the gravitational force on the planet when mass of the planet is twice of original mass is$8×{10}^{25}\mathrm{N}$

(c)

The distance between the star and the planet is given by,

$r^{\text{'}}=3r$

Here, $r$is the actual distance between the star and the planet.

The new gravitational force on the planet by the star is given by,

$\backslash begin\left\{aligned\right\}F\_\{PS\}^\{\backslash prime\}\&=\backslash frac\{Gm\_\{s\}m\_\{p\left\}\right\}\{r^\{\backslash prime2\left\}\right\}\backslash \backslash \&=\backslash frac\{Gm\_\{s\}m\_\{p\left\}\right\}\left\{\right(3r)^\{2\left\}\right\}\backslash \backslash \&=\backslash frac\{Gm\_\{s\}m\_\{P\left\}\right\}\{9r^\{2\left\}\right\}\backslash end\left\{aligned\right\}$

Here, $G$is a universal gravitational const ant value is $6.673×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2,m_P$is the mass of the planet, $m_s$is the mass of the star, and $r$is the distance between the center of the star and the planet.

Substitute this value in the force equation, the new force will be given by,

$\backslash begin\left\{aligned\right\}F\_\{PS\}^\{\backslash prime\}\&=\backslash frac\left\{1\right\}\left\{9\right\}F\_\{PS\}\backslash \backslash \&=\backslash frac\{4\backslash times10^\{25\}\backslash mathrm\{~N\left\}\right\}\left\{9\right\}\backslash \backslash \&=4.45\backslash times10^\left\{4\right\}\backslash mathrm\{~N\}\backslash end\left\{aligned\right\}$

Thus, the magnitude of the force when distance between the star and planet is trice the original distance is $4.45×{10}^4\mathrm{N}$.