A roughly spherical asteroid has a mass of $\mathbf{3}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{kg}$and a radius of 270 km. (a) What is the value of the constant g at a location on the surface of the asteroid? (b) What would be the magnitude of the gravitational force exerted by the asteroid on a 70 kg astronaut standing on the asteroid’s surface? (c) How does this compare to the gravitational force on the same astronaut when standing on the surface of the Earth?

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

The Radius of the asteroid is,$r=270\hspace{0.33em}km.$

The mass of astronaut is,$m=70\hspace{0.33em}kg.$

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 $6.673\times {10}^{-11}\hspace{0.33em}N\cdot m^2/k{g}^2,M_a$is the mass of the asteroid, r is the radius.

Substitute values in the above equation.

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

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.