\(a_c = \dfrac{v^{2}}{r}= \omega^{2}r = \dfrac{4\pi^{2}r}{T^{2}}\).
\[
v = \frac{2\pi r}{T}= r\omega ,\qquad
a_t = \frac{dv}{dt}\;( \text{tangential}),\qquad
a_c = \frac{v^{2}}{r}= \omega^{2}r = \frac{4\pi^{2}r}{T^{2}} .
\]
\[
a_c = \frac{v^{2}}{r}.
\]
\[
a_c = (\omega r)^{2}/r = \omega^{2}r .
\]
\[
a_c = \left(\frac{2\pi}{T}\right)^{2}r = \frac{4\pi^{2}r}{T^{2}} .
\]
All three expressions are mathematically identical; choose the one that contains the quantities given in the question.
| Source | Typical example | Force expression (inward) |
|---|---|---|
| Tension | Ball on a string | \(T = \dfrac{mv^{2}}{r}\) |
| Friction | Car on a flat curve | \(F_f = \mu N = \mu mg = \dfrac{mv^{2}}{r}\) |
| Normal reaction (banked curve) | Road banked at angle θ | \(N\sin\theta = \dfrac{mv^{2}}{r}\) with \(N\cos\theta = mg\) |
| Gravity | Satellite in orbit | \(\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}\) |
\[
\begin{aligned}
N\sin\theta &= \frac{mv^{2}}{r},\\
N\cos\theta &= mg .
\end{aligned}
\]
Eliminate \(N\):
\[
v = \sqrt{rg\tan\theta } .
\]
For \(r=50\;\text{m}\) and \(\theta =30^{\circ}\):
\[
v = \sqrt{50\times9.81\times\tan30^{\circ}} \approx 12.0\;\text{m s}^{-1}.
\]
The gravitational field \(\mathbf g\) at a point is the force per unit mass that a test mass would experience there:
\[
\mathbf g = \frac{\mathbf F_g}{m}\qquad\text{(units: m s⁻²)} .
\]
\[
g = \frac{GM}{r^{2}},\qquad \mathbf g \text{ directed toward the centre of mass}.
\]
Potential \(V\) is the work done per unit mass in bringing a test mass from infinity to a distance \(r\):
\[
V = -\frac{GM}{r}\qquad\text{(units: J kg⁻¹)} .
\]
The field and potential are related by \(\mathbf g = -\nabla V\).
\[
\begin{aligned}
R_E &= 6.37\times10^{6}\;\text{m},\quad h = 4.0\times10^{5}\;\text{m},\\
r &= R_E + h = 6.77\times10^{6}\;\text{m},\\[4pt]
g &= \frac{GM_E}{r^{2}} = \frac{3.986\times10^{14}}{(6.77\times10^{6})^{2}}
\approx 8.7\;\text{m s}^{-2}.
\end{aligned}
\]
\[
a_c = \frac{v^{2}}{r} = g = \frac{GM}{r^{2}} .
\]
\[
v = \sqrt{\frac{GM}{r}} .
\]
\[
\begin{aligned}
v &= \frac{2\pi r}{T}\;\Longrightarrow\;
T = \frac{2\pi r}{v}
= 2\pi\sqrt{\frac{r^{3}}{GM}} .
\end{aligned}
\]
A satellite is geostationary when its period equals the Earth’s rotation period (\(T=86400\;\text{s}\)).
Setting \(T\) in the period formula gives the required radius:
\[
r{\text{geo}} = \left(\frac{GME T^{2}}{4\pi^{2}}\right)^{\!1/3}
\approx 4.22\times10^{7}\;\text{m}.
\]
Altitude above the surface:
\[
h{\text{geo}} = r{\text{geo}}-R_E \approx 3.58\times10^{7}\;\text{m}
\;(35\,800\;\text{km}).
\]
Orbital speed at this radius:
\[
v{\text{geo}} = \frac{2\pi r{\text{geo}}}{T}
\approx 3.07\times10^{3}\;\text{m s}^{-1}.
\]
| Quantity | Symbol | Expression | Source / Physical origin | Units |
|---|---|---|---|---|
| Centripetal acceleration | \(a_c\) | \(\dfrac{v^{2}}{r}= \omega^{2}r = \dfrac{4\pi^{2}r}{T^{2}}\) | Resultant inward acceleration of any circular motion | m s⁻² |
| Gravitational field strength | g | \(\dfrac{GM}{r^{2}}\) | Mass \(M\) (Earth, planet, star) | m s⁻² |
| Centripetal force | \(F_c\) | \(\dfrac{mv^{2}}{r}=m\omega^{2}r\) | Provided by tension, friction, normal reaction, or gravity | N |
| Gravitational force | \(F_g\) | \(\dfrac{GMm}{r^{2}}\) | Interaction between two masses | N |
| Gravitational potential | V | \(-\dfrac{GM}{r}\) | Work per unit mass from infinity to \(r\) | J kg⁻¹ |
Speed \(v = 20\;\text{m s}^{-1}\), curve radius \(r = 50\;\text{m}\).
\[
a_c = \frac{v^{2}}{r}= \frac{20^{2}}{50}=8.0\;\text{m s}^{-2}.
\]
Force required for a 1500 kg car:
\[
Fc = mac = 1500\times8.0 = 1.2\times10^{4}\;\text{N}.
\]
\(g \approx 8.7\;\text{m s}^{-2}\).
Corresponding orbital speed:
\[
v = \sqrt{gr}= \sqrt{8.7\times6.77\times10^{6}} \approx 7.68\times10^{3}\;\text{m s}^{-1}.
\]
\(M_M = 6.42\times10^{23}\;\text{kg}\), orbital radius \(r = 1.0\times10^{7}\;\text{m}\).
\[
v = \sqrt{\frac{GM_M}{r}}
= \sqrt{\frac{6.67\times10^{-11}\times6.42\times10^{23}}{1.0\times10^{7}}}
\approx 3.28\times10^{3}\;\text{m s}^{-1}.
\]
\[
a_c = \frac{v^{2}}{r}\approx1.08\;\text{m s}^{-2},
\qquad g = \frac{GM_M}{r^{2}} \approx 1.08\;\text{m s}^{-2}.
\]
\(r = 80\;\text{m}\), banking angle \(\theta = 20^{\circ}\).
\[
v = \sqrt{rg\tan\theta}= \sqrt{80\times9.81\times\tan20^{\circ}}
\approx 13.2\;\text{m s}^{-1}.
\]
\[
r{\text{geo}} = \left(\frac{GME T^{2}}{4\pi^{2}}\right)^{1/3}
\approx 4.22\times10^{7}\;\text{m},
\]
\[
v{\text{geo}} = \frac{2\pi r{\text{geo}}}{T}
\approx 3.07\times10^{3}\;\text{m s}^{-1}.
\]
Show that the total mechanical energy per unit mass is \(\displaystyle \frac{E}{m}= -\frac{GM}{2r}\).
Solution:
Kinetic energy per unit mass \(= \frac{v^{2}}{2}= \frac{GM}{2r}\).
Potential energy per unit mass \(= -\frac{GM}{r}\).
Hence \(E/m = \frac{GM}{2r} - \frac{GM}{r}= -\frac{GM}{2r}\).
(This result is useful for exam questions on orbital energy.)
| Symbol | Definition | Formula |
|---|---|---|
| \(\theta\) | Angular displacement (rad) | — |
| \(\omega\) | Angular speed (rad s⁻¹) | \(\omega = \dfrac{2\pi}{T}=2\pi f\) |
| \(v\) | Linear (tangential) speed (m s⁻¹) | \(v = r\omega = \dfrac{2\pi r}{T}\) |
| \(a_c\) | Centripetal acceleration (m s⁻²) | \(\dfrac{v^{2}}{r}= \omega^{2}r = \dfrac{4\pi^{2}r}{T^{2}}\) |
| \(Fc\) | Centripetal force (N) | \(Fc = ma_c = \dfrac{mv^{2}}{r}\) |
| \(g\) | Gravitational field strength (m s⁻²) | \(g = \dfrac{GM}{r^{2}}\) |
| \(V\) | Gravitational potential (J kg⁻¹) | \(V = -\dfrac{GM}{r}\) |
| \(T\) | Period of revolution (s) | \(T = \dfrac{2\pi r}{v}=2\pi\sqrt{\dfrac{r^{3}}{GM}}\) |
| \(f\) | Frequency (Hz) | \(f = 1/T\) |
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