# The near-earth asteroid Rendezvous (NEAR), after traveling 2.1 billion km, is meant to orbit the...

## Question:

The near-earth asteroid Rendezvous (NEAR), after traveling 2.1 billion km, is meant to orbit the asteroid Eros at a height of about 15 km from the asteroidal center. Eros is roughly box-shaped, with dimensions 40 km {eq}\times {/eq} 6 km {eq}\times {/eq} 6 km. Assume eros has a density of about 2.3 {eq}\times {/eq} 10{eq}^3 {/eq} kg/m{eq}^3 {/eq}.

a) What will be the period of NEAR as it orbits Eros?

b) If eros were a sphere with the same mass and density, what would be its radius?

c) What would g be at the surface of this spherical Eros?

## Kepler's Law of Periods:

{eq}T^2=\frac{4\pi^2}{GM}a^3 {/eq}

Here {eq}T {/eq} is the period, {eq}a {/eq} is the semi-major axis of the orbit, {eq}M {/eq} is the mass of the host body, and the gravitational constant is {eq}G=6.674 \times 10^{-11} \frac{m^3}{kg \cdot s^2} {/eq}

For NEAR we assume a nearly circular orbit, such that the distance from the center of the asteroid is also the semi-major axis.

{eq}a=1.5 \times 10^4 m {/eq}

We can calculate Eros' mass from it's density and volume

{eq}M=(40000\cdot 6000 \cdot 6000)m^3(2.3\times 10^3)\frac{kg}{m^3}=3.312 \times 10^15 kg {/eq}

The orbital period is then:

{eq}T=\sqrt{\frac{4\pi^2}{GM}a^3}=\sqrt{\frac{4\pi^2}{(6.674 \times 10^{-11} \frac{m^3}{kg \cdot s^2})(3.312 \times 10^15 kg)}(1.5 \times 10^4 m)^3}=2.45 \times 10^4 s = \boxed{6.8 hr} {/eq}

Eros' volume is:

{eq}V=(40000\cdot 6000 \cdot 6000)m^3=1.44 \times 10^{12} m^3 {/eq}

The volume of a sphere is given by:

{eq}V=\frac{4}{3}\pi r^3 {/eq}

{eq}r=(\frac{3V}{4\pi})^{1/3}=(\frac{3(1.44 \times 10^{12} m^3)}{4\pi})^{1/3}=7000 m= \boxed{7.00 km} {/eq}

The gravitation acceleration for a specified mass at a given radius is:

{eq}g=\frac{GM}{r^2}=\frac{(6.674 \times 10^{-11} \frac{m^3}{kg \cdot s^2})(3.312 \times 10^15 kg)}{(7000 m)^2}=\boxed{4.50 \times 10^3 \frac{m}{s^2}} {/eq}