The Near Earth Asteriod Rendezvous (NEAR) spacecraft ,after traveling 2.1 billion km, is meant to orbit the asteroid Eros with an orbital radius of about 20 km. Eros is roughly \({\bf{40}}\;{\bf{km}} \times {\bf{6}}\;{\bf{km}} \times {\bf{6}}\;{\bf{km}}\). Assume Eros has a density (mass/volume) of about \({\bf{2}}{\bf{.3}} \times {\bf{1}}{{\bf{0}}^{\bf{3}}}\;{\bf{kg}}{\rm{/}}{{\bf{m}}^{\bf{3}}}\). (a) If Eros were a sphere with the same mass and density, what would its radius be? (b) What would g be at the surface of a spherical Eros? (c) Estimate the orbital period of NEAR as it orbits Eros, as if Eros were a sphere.

The mass of the Eros is given by the product of density and volume. Using the mass, the radius of Eros can be determined.

The acceleration due to gravity is inversely proportional to the square of radius of Eros.

The time period to revolve the satellite around the Earth is dependent on the velocity of Eros.

The orbital radius of Eros is \(20\;{\rm{km}}\).

The density of Eros is \(\rho = 2.3 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\)

The volume of the Eros is \(40\;{\rm{km}} \times {\rm{6}}\;{\rm{km}} \times {\rm{6}}\;{\rm{km}}\).

(a)

The mass of the Eros can be calculated as,

\(\begin{aligned}M &= \rho V\\ &= \left( {2.3 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {4000 \times 6000 \times 6000\;{{\rm{m}}^3}} \right)\\ &= 3.312 \times {10^{15}}\;{\rm{kg}}\end{aligned}\)

The radius of the Eros can be calculated as follows:

\(\begin{aligned}r &= {\left( {\frac{{3M}}{{4\pi \rho }}} \right)^{\frac{1}{3}}}\\ &= {\left( {\frac{{3\left( {3.312 \times {{10}^{15}}\;{\rm{kg}}} \right)}}{{4\pi \left( {2.3 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}} \right)}}} \right)^{\frac{1}{3}}}\\ &= 7005\;{\rm{m}}\\ \approx {\rm{7}} \times {\rm{1}}{{\rm{0}}^3}\;{\rm{m}}\end{aligned}\)

Thus, the radius of Eros is \(7 \times {10^3}\;{\rm{m}}\).

(b)

The acceleration g at the surface of Eros can be calculated as,

\(\begin{aligned}g &= \frac{{GM}}{{{r^2}}}\\ &= \frac{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {3.312 \times {{10}^{15}}\;{\rm{kg}}} \right)}}{{{{\left( {7005\;{\rm{m}}} \right)}^2}}}\\ &= 4.502 \times {10^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}\\ \approx 5 \times {10^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

Thus, the acceleration at the surface of Eros is \(5 \times {10^{ - 3}}\;{\rm{m/}}{{\rm{s}}^2}\)

(c)

The orbital velocity can be calculated as,

\(v = \sqrt {\frac{{GM}}{r}} \)

The orbital period can be calculated as,

\(\begin{aligned}T &= \frac{{2\pi r}}{v}\\ &= \frac{{2\pi r}}{{\sqrt {\frac{{GM}}{r}} }}\end{aligned}\)

Substitiue the values in the above equation,

\(\begin{aligned}T &= \frac{{2\pi \left( {2 \times {{10}^4}\;{\rm{m}}} \right)}}{{\sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {3.312 \times {{10}^{15}}\;{\rm{kg}}} \right)}}{{\left( {2 \times {{10}^4}\;{\rm{m}}} \right)}}} }}\\ &= 3.781 \times {10^4}\;{\rm{s}}\left( {\frac{{1\;{\rm{hr}}}}{{3600\;{\rm{s}}}}} \right)\\ \approx 11\;\;{\rm{h}}\end{aligned}\)

Thus, the orbital period of Eros is \(11\;{\rm{h}}\).