91Ó°ÊÓ

Mass of star equal to mass of sun

Radius of star is $r=10.0km  or  10000 m$

Distance of object is 1.0 m

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

With the mass and radius of a star, we can find its gravitational acceleration. Using the law of energy conservation, we can find its velocity.

Formula:

$a_g=\frac{GM}{r^2}$ …(¾±)

$K=\frac{1}{2}×M×V^2$ …(¾±¾±)

$U=-\frac{GMm}{r}$ …(¾±¾±¾±)

Where, a_g is gravitational acceleration

G is the gravitational constant ( $6.67×{10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

M is the mass of star = mass of sun(role="math" $2.0×{10}^{30} \text{kg}$)

K is the kinetic energy

U is the potential energy

V is the speed of object

From equation (i)

$$\begin{gathered} a_g=\frac{GM}{r^2} \\ =\frac{\left(6.67×{10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}\right)\left(2.0×{10}^{30} \text{kg}\right)}{{\left(10000 \text{m}\right)}^2} \\ =1.33×{10}^{12} {\text{m/s}}^{\text{2}} \end{gathered}$$

By using energy conservation law,

$K{E}_{initial}+P{E}_{initial}=K{E}_{final}+P{E}_{final}$

From equation (ii) and equation (iii)

$$\begin{gathered} 0+\frac{GMm}{r_0}=\frac{1}{2}m{v}^2+\frac{GMm}{r} \\ \frac{1}{2}{v}^2=\frac{GM}{r}-\frac{GM}{r_0} \\ {v}^2=2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)} \end{gathered}$$

After plugging the values, we get $r_0=10001 \text{m}$

$$\begin{gathered} v=\sqrt{2\left(6.67×{10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}\right)\left(2.0×{10}^{30} \text{kg}\right)\left(\frac{\text{1}}{10000 \text{m}}-\frac{1}{10001 \text{m}}\right)} \\ v=1.62×{10}^6 \text{m/s} \end{gathered}$$

Hence, the speed of object is $1.62×{10}^6 \text{m/s}$