NESA Physics Circular Motion

The satellite has inertia and, in the absence of the planet, would continue in a straight line with the same speed until acted upon by an unbalanced force.

When the planet's gravitational force pulls the satellite towards it (perpendicular to the satellite's motion), the satellite changes direction and thus accelerates towards the planet, but its speed does not change.

As a result, the satellite continues to move ‘forward’, but the planet’s gravitational force pulls the satellite towards it. So the resultant motion is that the satellite has a constant speed (not velocity as direction is changing) travelling in a circle around the planet.

The centripetal force is supplied by the reaction force, so $mv^2/r = R$.

The reaction force equals $mg$, giving $mv^2/r = mg$.

Rearranging the formula to calculate velocity gives $v = \sqrt{rg} = \sqrt{4.60 \times 10^2 \times 9.80} = 67.1 \text{ m s}^{-1}$.

The period is circumference over time, $T = 2\pi r/v$.

Calculating the period gives $T = 43.0 \text{ s}$.

$C = 2\pi r = 2 \times \pi \times 35 \approx 219.91 \text{ m}$

$f = \frac{1}{T}$

$\therefore v = \frac{2\pi r}{T} = 219.91 \times 1.3 \times 10^6 = 2.86 \times 10^8 \text{ m s}^{-1}$

Average speed $= 2.9 \times 10^8 \text{ m s}^{-1}$ (to two significant figures)

$F = \frac{mv^2}{r}$
Let $R$ be the radius of the new path
$\frac{Mv^2}{r} = \frac{2Mv^2}{R}$
$\frac{1}{r} = \frac{2}{R}$
$R = 2r$
The radius will double.

distance = $2\pi r \times$ number of revolutions
$= 2\pi \times 0.25 \times \frac{3.9}{0.3}$
$= 20.4 m$

Distance travelled = 20 m (to two significant figures)

$v = \frac{2\pi r}{T}$
$= \frac{2\pi \times 0.25}{0.3}$
$= 5.236... \text{ m s}^{-1}$
$F_{net} = \frac{mv^2}{r}$
$= \frac{0.2 \times 5.236^2}{0.25}$
$= 21.93 N$

Centripetal force = 22 N (to two significant figures)

$v = \frac{2\pi r}{T}$
$= \frac{2\pi \times 0.8}{5 \div 12}$
$= 12.1\ m\ s^{-1}$

$a_c = \frac{v^2}{r}$
$= \frac{12.1^2}{0.8}$
$= 180\ m\ s^{-2}$

Centripetal acceleration = 180 m s$^{-2}$ (to two significant figures)

$s_y = u_y t + \frac{1}{2} a t^2$
$4.3 = 0 + \frac{1}{2} \times 9.8 t^2$
$t = \sqrt{\frac{4.3}{4.9}}$
$= 0.94\ s$

$s_x = u_x t + \frac{1}{2} a t^2$
$= 12.1 \times 0.94 + \frac{1}{2} \times 0 \times 0.94^2$
$= 11.4\ m$

Horizontal displacement = 11 m (to two significant figures)