In June 1997 the NEAR spacecraft , on its way to photograph the asteroid Eros, passed near the asteroid Mathilde. After passing Mathilde, on several occasions rocket propellant was expelled to adjust the spacecraft's momentum in order to follow a path that would approach the asteroid Eros, the final destination for the mission. After getting close to Eros, further small adjustments made the momentum just right to give a circular orbit of radius around the asteroid. So much propellant had been used that the final mass of the spacecraft while in circular orbit $\mathbf{45}\mathbf{\text{km}}\left(45\times {10}^3\text{m}\right)$around Eros was only . The spacecraft took $\mathbf{1}\mathbf{.}\mathbf{04}$days to make one complete circular orbit around Eros. Calculate what the mass of Eros must be.

A circular orbit of radius is$45\text{km}\left(45\times {10}^3\text{m}\right)$.

The final mass of the spacecraft while in circular orbit is$500\text{kg}$.

The spacecraft took days to make one complete circular orbit around Eros.

A force is a push or pull on an object is because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them. The acted force may be of attraction or repulsion. The two items no longer feel the force after the interaction ends.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}$and the perpendicular rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\perp }$are the two elements that we are concerned with.

So, the net force$F_{net}$on the object is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ ={\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}+{\frac{d\overrightarrow{p}}{dt}}_{\perp } \end{gathered}$$

The rock's speed is affected by the parallel rate of change of momentum and given the speed is constant, therefore, the parallel rate is zero, and it equals the size of the momentum rate change.

${\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}=0$

The centrifugal force$F_c$equals the rate change, which is the change in direction owing to the perpendicular rate of change.

At speeds significantly slower than the speed of light, the magnitude of the perpendicular rate change is given by

$$\begin{gathered} F_c={\frac{d\overrightarrow{p}}{dt}}_{\perp } \\ =\frac{m{v}^2}{R} \end{gathered}$$

The change in distance over time is the speed.

So, it is given by

$v=\frac{d}{t}$

The engineer goes above the diameter of a circle as he travels through one cycle.

The circumference is given by,

$2\pi R$

where R is the radius of the circle.

So, the distance where the engineer travel is:

$d=2\pi R$

Therefore,

$v=\frac{2\pi R}{t}$

The mass of the Eros,

$$\begin{gathered} v_{earth}=v_{asteroid} \\ \sqrt{\frac{G{M}_E}{R}}=\frac{2\pi R}{t} \\ \frac{G{M}_E}{R}={\left(\frac{2\pi R}{t}\right)}^2 \\ {M}_E={\left(\frac{2\pi R}{t}\right)}^2\left(\frac{R}{G}\right) \end{gathered}$$

Now, put the values for $R,t,andGtoget{M}_E$, where is the time taken to complete one round by the spacecraft.

$$\begin{gathered} t=(1.04\text{days})(24\text{h}/\text{day})(3600\text{s}/\text{h}) \\ =89.85\times {10}^3\text{s} \end{gathered}$$

$$\begin{gathered} M_E={\left(\frac{2\pi R}{t}\right)}^2\left(\frac{R}{G}\right)\frac{1}{n} \\ ={\left(\frac{2\pi \times 45\times {10}^3\text{m}}{89.85\times {10}^3\text{s}}\right)}^2\left(\frac{45\times {10}^3\text{m}}{6.67\times {10}^{-11}\text{N}·{\text{m}}^2/{\text{kg}}^2}\right) \\ =6.67\times {10}^{15}\text{kg} \end{gathered}$$

Therefore, the mass of the Eros is $6.67\times {10}^{15}\text{kg}$.