91Ó°ÊÓ

The mass of the asteroid is$m_a=3.11×{10}^{20}\mathrm{kg}$

The Radius of the asteroid is,$r=270\mathrm{km}$

The mass of astronaut is,$m=70\mathrm{kg}$

Gravity is a force that pulls items together over a long distance by acting on their mass.

Gravity is the curvature of spacetime driven by mass energy, which governs mass-energy paths (equations of motion).

(a)

Acceleration due to gravity on asteroid is,

$g=\frac{G{m}_{\mathrm{a}}}{r^2}$

Here, $G$is the universal gravitational constant whose value is given by $6.673×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2,m_{\mathrm{s}}$is the mass of the asteroid, $r$is the radius.

Substitute values in the above equation.

$\backslash begin\left\{aligned\right\}g\&=\backslash frac\{6.673\backslash times10^\{-11\}\backslash mathrm\{~N\}\backslash cdot\backslash mathrm\{m\}^\{2\}/\backslash mathrm\{kg\}^\{2\}\backslash left(3.11\backslash times10^\left\{20\right\}\backslash mathrm\{~kg\}\backslash right\left)\right\}\left\{\right(270\backslash mathrm\{~km\})^\{2\left\}\right\}\backslash \backslash \&=\backslash frac\{6.673\backslash times10^\{-11\}\backslash mathrm\{~N\}\backslash cdot\backslash mathrm\{m\}^\{2\}/\backslash mathrm\{kg\}^\{2\}\backslash left(3.1\backslash times10^\left\{20\right\}\backslash mathrm\{~kg\}\backslash right\left)\right\}\{\backslash left[270\backslash mathrm\{~km\}\backslash left(\backslash frac\{1000\backslash mathrm\{~m\}\left\}\right\{1\backslash mathrm\{~km\}\}\backslash right)\backslash right]^\{2\left\}\right\}\backslash \backslash \&=0.284\backslash mathrm\{~N\}/\backslash mathrm\left\{kg\right\}\backslash end\left\{aligned\right\}$

Thus, the value of the $g$at a location on the surface of the asteroid is $284\backslash mathrm\{~N\}/\backslash mathrm\left\{kg\right\}$

(b)

The gravitational force on an astronaut of mass $m$is given by.

$F=mg$

Here, $m$is the mass, and $g$is the acceleration due to gravity.

Substitute values in the above equation.

$\backslash begin\left\{aligned\right\}F\&=70\backslash mathrm\{~kg\}(0.284\backslash mathrm\{~N\}/\backslash mathrm\{kg\left\}\right)\backslash \backslash \&=19.88\backslash mathrm\{~N\}\backslash end\left\{aligned\right\}$

Thus, the force on astronaut is$19.88\mathrm{N}$

(c)

The gravitational force on the astronaut is given by,

$F^{\text{'}}=m{g}^{\text{'}}$

Here, $m$is the mass and $g$is the acceleration due to gravity on the surface of the earth.

Substitute values in the above equation.

$\backslash begin\left\{aligned\right\}F\&=70\backslash mathrm\{~kg\}(9.8\backslash mathrm\{~N\}/\backslash mathrm\{kg\left\}\right)\backslash \backslash \&=686\backslash mathrm\{~N\}\backslash end\left\{aligned\right\}$

Thus, the result states that the gravitational force on asteroid is very small compared to that on surface of the earth.