91Ó°ÊÓ

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

The Radius of the asteroid is,$r=270 km.$

The mass of astronaut is,$m=70 kg.$

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,

$g=\frac{G{m}_a}{r^2}$

Here, G is the universal gravitational constant whose value is given by $6.673×{10}^{-11} Nâ‹\dots m^2/k{g}^2,M_a$is the mass of the asteroid, r is the radius.

Substitute values in the above equation.

localid="1658070470018" $$\begin{gathered} g=\frac{6.673×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\left(3.11×{10}^{20}\mathrm{kg}\right)}{(270\mathrm{km}{)}^2} \\ =\frac{6.673×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\left(3.1×{10}^{20}\mathrm{kg}\right)}{{\left[270\mathrm{km}\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\right]}^2} \\ =0.284N/kg. \end{gathered}$$

Thus, the value of the gat a location on the surface of the asteroid is $0.284N/kg.$

(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.

$$\begin{gathered} F=70\mathrm{kg}(0.284\mathrm{N}/\mathrm{kg}) \\ =19.88N \end{gathered}$$

Thus, the force on astronaut is $19.88N$.

(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.

$$\begin{gathered} F=70\mathrm{kg}(9.8\mathrm{N}/\mathrm{kg}) \\ =686N \end{gathered}$$

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