91Ó°ÊÓ

The radius of the circular path$r=3220 \text{km}=3.220×{10}^6 \text{m}$

The speed of the spaceship$v=29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.$

The spaceship is taking a turn along a circular path. So, we need to use the equations that relates the linear variables (v and a) with the corresponding angular variables ($\mathrm{Ó\neg }$and$\mathrm{Î\pm }$). The two variables are related through the parameter r, the radius of the path.

Formula:

$v=r\mathrm{Ó\neg }$

$a_r=\frac{v^2}{r}$

$a_t=\mathrm{Î\pm }r$

We convert the speed of the spaceship to the units of m/s

$\begin{array}{rcl}v& =& 29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.\\ & =& 29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.×\frac{1000 \text{m}}{1 \text{km}}×\frac{1 \text{hr}}{3600 \text{s}}\\ & =& 8.05×{10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The angular velocity and the linear velocity are related as

$\mathrm{Ó\neg }=\frac{v}{r}$

Substitute all the value in the above equation.

role="math" localid="1660903601326" $$\begin{gathered} \mathrm{Ó\neg }=\frac{{\displaystyle 8.05×{10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{{\displaystyle 3.22×{10}^{6}m}} \\ \mathrm{Ó\neg }=2.5×{10}^{-3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The magnitude of the angular velocity of the spaceship is$2.5×{10}^{-3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The radial acceleration is calculated as

$a_r=\frac{v^2}{r}$

Substitute all the value in the above equation.

role="math" localid="1660903859409" $$\begin{gathered} a_r=\frac{{\displaystyle {\left(8.05×{10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2}}{{\displaystyle 3.22×{10}^6 \text{m}}} \\ {a}_r=20.2\raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right. \end{gathered}$$

The magnitude of the radial acceleration of the spaceship is $20.2\raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right.$.

The tangential acceleration can be calculated as$a_t=\mathrm{Î\pm }r$where$\mathrm{Î\pm }$is the angular acceleration. This is related to the change in the magnitude of the linear velocity v. The given spaceship is taking a turn at constant linear velocity. Hence, the magnitude of the velocity is constant. So, the angular accelerationis zero. Hence, the tangential component of the linear acceleration is also zero.

$$\begin{gathered} a_t=\mathrm{Î\pm }r \\ {a}_t=0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The magnitude of the tangential acceleration of the spaceship is,$0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .