Concept Card: Motion of a Satellite – Orbital Velocity & Time Period
1. Mechanics of Satellite Motion:
A satellite is any natural or artificial body describing a stable orbit around a more massive primary body under the exclusive action of mutual gravitational attraction. For a satellite of mass $m$ describing a circular orbit of radius $r$ centered on a planet of mass $M$ and radius $R$ at an altitude $h$ ($r = R + h$):
The necessary centripetal force required to maintain circular motion is supplied entirely by the gravitational attraction exerted by the primary body:
$$F_{\text{centripetal}} = F_{\text{gravitational}}$$
$$\frac{m v_o^2}{r} = \frac{G M m}{r^2}$$
where $v_o$ is the orbital velocity of the satellite.
2. Rigorous Derivation of Orbital Velocity ($v_o$)
Canceling the satellite mass $m$ and one power of $r$ from both sides:
$$v_o^2 = \frac{G M}{r} \implies \mathbf{v_o = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{R + h}}}$$
Alternative Formulations:
- In terms of Surface Acceleration Due to Gravity ($g_0 = \frac{G M}{R^2} \implies G M = g_0 R^2$):
$$\mathbf{v_o = \sqrt{\frac{g_0 R^2}{R + h}} = R \sqrt{\frac{g_0}{R + h}}}$$ - Near the Surface of the Earth ($h \ll R \implies r \approx R$):
$$\mathbf{v_{o, \text{surface}} = \sqrt{\frac{G M}{R}} = \sqrt{g_0 R}}$$
Numerical Evaluation for Earth: With $R = 6.4 \times 10^6\text{ m}$ and $g_0 = 9.8\text{ m/s}^2$:
$$v_{o, \text{surface}} = \sqrt{9.8 \times 6.4 \times 10^6} = \sqrt{6.272 \times 10^7} \approx \mathbf{7.92\text{ km/s}}$$ - In terms of Earth’s Mean Mass Density ($\rho$):
$$M = \frac{4}{3}\pi R^3 \rho \implies v_{o, \text{surface}} = \sqrt{\frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R}} = \mathbf{R \sqrt{\frac{4}{3}\pi G \rho}}$$
Fundamental Relationship Between Orbital Velocity and Escape Velocity:
The escape velocity from the planet’s surface is $v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g_0 R}$. Comparing with surface orbital speed:
$$\mathbf{v_e = \sqrt{2} v_{o, \text{surface}} \approx 1.414 v_o} \quad \Longleftrightarrow \quad \mathbf{v_o = \frac{v_e}{\sqrt{2}} \approx 0.707 v_e}$$
Key Physical Principle:
If a satellite orbiting near the Earth’s surface has its velocity boosted by a factor of $\sqrt{2}$ (an increase of $\Delta v = (\sqrt{2} – 1) v_o \approx 41.4\%$), its closed circular bound orbit transforms into an open parabolic escape trajectory, and it permanently leaves the Earth’s gravitational field!
3. Rigorous Derivation of Orbital Time Period ($T$)
The time period $T$ is the total time taken by the satellite to complete one full revolution ($2\pi\text{ radians}$) around the planet:
$$T = \frac{\text{Circumference of orbit}}{\text{Orbital velocity}} = \frac{2\pi r}{v_o} = \frac{2\pi r}{\sqrt{\frac{G M}{r}}} = \frac{2\pi r \cdot r^{1/2}}{\sqrt{G M}}$$
Governing Equation for Time Period ($T$):
$$\mathbf{T = 2\pi \sqrt{\frac{r^3}{G M}} = 2\pi \sqrt{\frac{(R + h)^3}{G M}} = \frac{2\pi}{R} \sqrt{\frac{(R + h)^3}{g_0}}}$$
Special Cases & The Density Invariance Theorem:
- Near Earth’s Surface ($h \ll R$):
$$T_{\text{surface}} = 2\pi \sqrt{\frac{R^3}{g_0 R^2}} = 2\pi \sqrt{\frac{R}{g_0}}$$
Numerical Evaluation:
$$T_{\text{surface}} = 2\pi \sqrt{\frac{6.4 \times 10^6\text{ m}}{9.8\text{ m/s}^2}} \approx 2\pi (808.1\text{ s}) \approx \mathbf{5077\text{ seconds} \approx 84.6\text{ minutes} \approx 1.41\text{ hours}}$$
Note: This period is mathematically identical to the oscillation period of a particle dropped through a frictionless diametrical tunnel in the Earth! - The Density Invariance Theorem:
Substitute $M = \frac{4}{3}\pi R^3 \rho$ into the grazing surface period formula:
$$T_{\text{surface}} = 2\pi \sqrt{\frac{R^3}{G \left(\frac{4}{3}\pi R^3 \rho\right)}} = 2\pi \sqrt{\frac{3}{4\pi G \rho}} = \mathbf{\sqrt{\frac{3\pi}{G \rho}}}$$
Remarkable Law: The minimum orbital period of a surface-grazing satellite depends exclusively on the mean mass density $\rho$ of the celestial body, and is completely independent of its radius $R$ or mass $M$!
4. Angular Momentum, Areal Velocity & Parameter Scaling
- Orbital Angular Momentum ($L$):
$$L = m v_o r = m \left(\sqrt{\frac{G M}{r}}\right) r = \mathbf{m \sqrt{G M r} = m \sqrt{G M (R + h)}}$$
Angular momentum scales as $\sqrt{r}$. - Areal Velocity ($\frac{dA}{dt}$):
$$\frac{dA}{dt} = \frac{L}{2m} = \mathbf{\frac{1}{2} \sqrt{G M r} = \text{constant}}$$ - Orbital Angular Velocity ($\omega$):
$$\omega = \frac{v_o}{r} = \frac{\sqrt{G M / r}}{r} = \mathbf{\sqrt{\frac{G M}{r^3}}}$$
Fractional / Differential Variations for Small Radial Changes ($\frac{\Delta r}{r} \ll 1$):
- Since $v_o \propto r^{-1/2} \implies \mathbf{\frac{\Delta v_o}{v_o} = -\frac{1}{2} \frac{\Delta r}{r}}$ (speed decreases as radius increases).
- Since $T \propto r^{3/2} \implies \mathbf{\frac{\Delta T}{T} = +\frac{3}{2} \frac{\Delta r}{r}}$ (period increases as radius increases).
- Since $L \propto r^{1/2} \implies \mathbf{\frac{\Delta L}{L} = +\frac{1}{2} \frac{\Delta r}{r}}$.
- Since $\omega \propto r^{-3/2} \implies \mathbf{\frac{\Delta \omega}{\omega} = -\frac{3}{2} \frac{\Delta r}{r}}$.
5. The Atmospheric Drag Paradox (Satellite Acceleration Paradox)
When a low Earth satellite encounters slight atmospheric drag ($F_{\text{drag}}$):
- The drag force dissipates mechanical energy: $\Delta E < 0$.
- Because total mechanical energy in a circular orbit is $E = -\frac{G M m}{2r}$, a decrease in energy requires the orbital radius $r$ to decrease (the orbit slowly spirals inward).
- Because orbital velocity is $v_o = \sqrt{G M / r}$, as $r$ decreases, the orbital speed increases!
Conclusion: Atmospheric friction causes an orbiting satellite to speed up, not slow down! Gravitational potential energy decreases twice as fast as kinetic energy increases ($\Delta K = -\Delta E = |W_{\text{drag}}|$), transferring the difference into kinetic energy.
Solved Examples
Example 1 (Direct Ratio Application – Altitude Scaling of Velocity, Period & Momentum):
Satellite $S_1$ orbits Earth at an altitude $h_1 = R$, and satellite $S_2$ of identical mass orbits at an altitude $h_2 = 7R$, where $R$ is the Earth’s radius.
(a) Find the orbital radii $r_1$ and $r_2$ of the two satellites.
(b) Determine the ratio of their orbital speeds ($v_1 / v_2$).
(c) Determine the ratio of their orbital time periods ($T_1 / T_2$).
(d) Find the ratio of their orbital angular momenta ($L_1 / L_2$).
Solution:
(a) Orbital Radii:
$$r_1 = R + h_1 = R + R = 2R$$
$$r_2 = R + h_2 = R + 7R = 8R$$
(b) Ratio of Orbital Speeds:
Since $v_o = \sqrt{\frac{G M}{r}} \propto \frac{1}{\sqrt{r}}$:
$$\frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{8R}{2R}} = \sqrt{4} = \mathbf{2.0}$$
The lower satellite orbits twice as fast as the higher satellite.
(c) Ratio of Orbital Periods:
By Kepler’s Third Law, $T \propto r^{3/2}$:
$$\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{2R}{8R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8} = \mathbf{0.125}$$
$$\frac{T_2}{T_1} = 8.0$$
Satellite $S_2$ takes 8 times longer to complete one revolution.
(d) Ratio of Angular Momenta:
Since $L = m \sqrt{G M r} \propto \sqrt{r}$:
$$\frac{L_1}{L_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{2R}{8R}} = \sqrt{\frac{1}{4}} = \mathbf{\frac{1}{2} = 0.50}$$
Satellite $S_2$ carries twice the angular momentum of satellite $S_1$.
Example 2 (The Density Invariance Theorem – Planetary Comparative Analysis):
Two spherical planets $A$ and $B$ have radii $R_A = 2 R_B$ and identical uniform mean mass densities ($\rho_A = \rho_B$).
(a) Determine the ratio of the surface orbital speeds of satellites orbiting just above the surfaces of planets $A$ and $B$ ($v_{oA} / v_{oB}$).
(b) Determine the ratio of the orbital time periods of these surface satellites ($T_A / T_B$).
(c) Given $G = 6.674 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$ and $\rho = 5500\text{ kg/m}^3$, compute the numerical orbital period.
Solution:
(a) Surface Orbital Velocity:
For a grazing surface satellite:
$$v_o = R \sqrt{\frac{4}{3}\pi G \rho}$$
Since $\rho_A = \rho_B$:
$$\frac{v_{oA}}{v_{oB}} = \frac{R_A}{R_B} = \mathbf{2.0}$$
(b) Orbital Period:
By the density invariance theorem:
$$T = \sqrt{\frac{3\pi}{G \rho}}$$
Because $T$ depends only on density $\rho$, and $\rho_A = \rho_B$:
$$\frac{T_A}{T_B} = \mathbf{1.0}$$
The orbital periods are strictly identical regardless of the planetary radius!
(c) Numerical Computation:
$$T = \sqrt{\frac{3\pi}{(6.674 \times 10^{-11})(5500)}} = \sqrt{\frac{9.4248}{3.6707 \times 10^{-7}}} = \sqrt{2.5676 \times 10^7} \approx \mathbf{5067\text{ seconds} \approx 84.45\text{ minutes}}$$
Example 3 (Calculus & Perturbation – Small Orbit Adjustment Maneuvers):
A low Earth remote-sensing satellite circles Earth in a circular orbit of radius $r = 7000\text{ km}$ with an orbital period $T = 90.0\text{ minutes}$. An orbital adjustment thruster increases the orbital radius by $\Delta r = 14\text{ km}$.
(a) Determine the fractional and percentage change in the orbital radius $\Delta r / r$.
(b) Calculate the percentage change in the satellite’s orbital speed.
(c) Calculate the percentage change in the satellite’s orbital period.
(d) Find the time difference $\Delta T$ (in seconds) per revolution after the maneuver.
Solution:
(a) Fractional and Percentage Change in Radius:
$$\frac{\Delta r}{r} = \frac{14\text{ km}}{7000\text{ km}} = \frac{2}{1000} = 0.002 = \mathbf{+0.20\%}$$
(b) Percentage Change in Orbital Speed:
Differentiating $v_o = \sqrt{G M} r^{-1/2}$:
$$\frac{\Delta v_o}{v_o} = -\frac{1}{2} \frac{\Delta r}{r} = -\frac{1}{2}(+0.20\%) = \mathbf{-0.10\%}$$
The satellite slows down by $0.10\%$.
(c) Percentage Change in Orbital Period:
Differentiating $T = \frac{2\pi}{\sqrt{G M}} r^{3/2}$:
$$\frac{\Delta T}{T} = +\frac{3}{2} \frac{\Delta r}{r} = +\frac{3}{2}(+0.20\%) = \mathbf{+0.30\%}$$
The orbital period increases by $0.30\%$.
(d) Time Difference per Orbit:
$$\Delta T = (+0.003) T = (+0.003)(90.0 \times 60\text{ s}) = (+0.003)(5400\text{ s}) = \mathbf{+16.2\text{ seconds}}$$
Each orbit takes $16.2\text{ seconds}$ longer.
Example 4 (Theoretical Mechanics – Apparent Weightlessness Inside Orbiting Satellites):
An astronaut of mass $m = 80\text{ kg}$ is aboard a space station in circular orbit at an altitude $h = R/3$ above Earth’s surface.
(a) Derive the local acceleration due to gravity $g(h)$ in terms of surface gravity $g_0$.
(b) Calculate the true gravitational force exerted by Earth on the astronaut.
(c) Explain using Newton’s laws why a weighing scale inside the space station reads zero (apparent weightlessness).
Solution:
(a) Local Gravity at Altitude $h = R/3$:
$$r = R + \frac{R}{3} = \frac{4}{3} R$$
$$g(h) = \frac{G M}{r^2} = \frac{G M}{\left(\frac{4}{3} R\right)^2} = \frac{9}{16} \frac{G M}{R^2} = \mathbf{\frac{9}{16} g_0}$$
(b) True Gravitational Force:
$$W_{\text{true}} = m g(h) = m \left(\frac{9}{16} g_0\right) = 80 \times \left(\frac{9}{16} \times 9.8\right) = 80 \times 5.5125 = \mathbf{441.0\text{ N}}$$
(The astronaut’s true weight is $56.25\%$ of their surface weight, definitely not zero!).
(c) Apparent Weightlessness Explained:
Consider the equation of motion of the astronaut inside the spacecraft:
$$F_{\text{net}} = m a_c$$
The forces acting on the astronaut in the radial direction are the true downward gravitational force $F_g = m g(h)$ and the upward normal contact force $N$ exerted by the floor/scale:
$$m g(h) – N = m a_c$$
For circular orbit, the centripetal acceleration is:
$$a_c = \frac{v_o^2}{r} = \frac{G M / r}{r} = \frac{G M}{r^2} = g(h)$$
Substitute $a_c = g(h)$:
$$m g(h) – N = m g(h) \implies \mathbf{N = 0}$$
Because both the astronaut and the spacecraft are in continuous, simultaneous free fall toward Earth with the exact same acceleration $g(h)$, no contact reaction force develops between them ($N = 0$).
Practice Worksheet: Motion of a Satellite – Orbital Velocity & Time Period
Section A: Single Correct Questions (Q1 to Q5)
Q1. The orbital velocity of a satellite revolving close to the surface of the Earth is approximately $7.92\text{ km/s}$. What is the escape velocity from the same location?
(A) $7.92\text{ km/s}$
(B) $11.2\text{ km/s}$
(C) $15.84\text{ km/s}$
(D) $5.60\text{ km/s}$
Q2. What minimum percentage increase in the speed of a low-orbit Earth satellite is required for it to overcome Earth’s gravitational pull and escape to infinity?
(A) $20.7\%$
(B) $41.4\%$
(C) $50.0\%$
(D) $100.0\%$
Q3. A satellite is revolving in a circular orbit at an altitude $h = 3R$ above the Earth’s surface (where $R$ is Earth’s radius). If $v_s$ is the surface orbital velocity, the orbital velocity of this satellite is:
(A) $\frac{v_s}{4}$
(B) $\frac{v_s}{2}$
(C) $\frac{v_s}{\sqrt{2}}$
(D) $\frac{v_s}{3}$
Q4. Two satellites $S_1$ and $S_2$ revolve around Earth in circular orbits of radii $r_1 = R$ and $r_2 = 4R$ respectively. The ratio of their orbital time periods $T_1 : T_2$ is:
(A) $1 : 2$
(B) $1 : 4$
(C) $1 : 8$
(D) $1 : 16$
Q5. The minimum orbital period of an artificial satellite orbiting close to the surface of any spherical planet of radius $R$ and uniform density $\rho$ depends exclusively on:
(A) The radius $R$ of the planet
(B) The mass $M$ of the planet
(C) The mean density $\rho$ of the planet
(D) The mass $m$ of the satellite
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. For a satellite revolving in a stable circular orbit of radius $r$ around Earth, which of the following statements is/are correct?
(A) The orbital speed is inversely proportional to $\sqrt{r}$.
(B) The orbital period is directly proportional to $r^{3/2}$.
(C) The orbital angular momentum about the center of Earth is directly proportional to $\sqrt{r}$.
(D) The centripetal acceleration is inversely proportional to $r^2$.
Q7. A satellite in low Earth orbit experiences a very small resistive drag force due to the upper tenuous atmosphere. Which of the following statements is/are correct regarding its subsequent motion?
(A) The total mechanical energy of the satellite decreases over time.
(B) The orbital radius $r$ of the satellite decreases.
(C) The linear orbital speed $v_o$ of the satellite increases.
(D) The kinetic energy of the satellite increases.
Q8. An astronaut inside an orbiting space station experiences weightlessness because:
(A) The gravitational force of the Earth on the astronaut is zero.
(B) The space station and the astronaut have identical accelerations toward the Earth’s center.
(C) The normal contact reaction between the astronaut and the space station floor is zero.
(D) The centrifugal force in the orbiting frame exactly balances the gravitational force.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A satellite orbits the Earth in a circular orbit at an altitude $h = R$ above the Earth’s surface, where $R$ is the Earth’s radius. The orbital speed of this satellite can be expressed as $v = \sqrt{\frac{g_0 R}{n}}$, where $g_0$ is the surface acceleration due to gravity. Find the integer value of $n$.
Q10. Two satellites $A$ and $B$ are in circular orbits around Earth with orbital radii $r_A = R$ and $r_B = 9R$ respectively. The ratio of their orbital time periods $\frac{T_B}{T_A}$ is an integer. Find this integer.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (B) | Escape vs Orbital Speed: $v_e = \sqrt{2} v_o \approx 11.2\text{ km/s}$ |
| Q2 | (B) | Escape Boost Fraction: $(\sqrt{2} – 1) \times 100\% \approx 41.4\%$ |
| Q3 | (B) | Radius Scaling: $r = 4R \implies v_o = \frac{1}{\sqrt{4}} v_s = \frac{v_s}{2}$ |
| Q4 | (C) | Kepler’s Third Law: $T_1/T_2 = (1/4)^{3/2} = 1/8$ |
| Q5 | (C) | Density Invariance Theorem: $T = \sqrt{3\pi / (G \rho)}$ |
| Q6 | (A, B, C, D) | Complete Scaling Laws of Circular Satellite Orbits |
| Q7 | (A, B, C, D) | The Satellite Atmospheric Drag Paradox Dynamics |
| Q8 | (B, C, D) | Mechanics of Apparent Weightlessness in Orbit |
| Q9 | 2 | Orbital Speed at $r = 2R$: $v = \sqrt{g_0 R / 2} \implies n = 2$ |
| Q10 | 27 | Period Ratio: $(9)^{3/2} = 27$ |
Detailed Explanations
Solution 1:
The relation between surface escape velocity and surface orbital velocity is:
$$v_e = \sqrt{2} v_o = \sqrt{2} (7.92\text{ km/s}) \approx 1.414 \times 7.92 = \mathbf{11.2\text{ km/s}}$$
Correct Answer: (B)
Solution 2:
To escape from a circular orbit, the satellite must reach escape speed $v_e = \sqrt{2} v_o$.
The required percentage increase in velocity is:
$$\% \text{ increase} = \frac{v_e – v_o}{v_o} \times 100\% = (\sqrt{2} – 1) \times 100\% = (1.414 – 1) \times 100\% = \mathbf{41.4\%}$$
Correct Answer: (B)
Solution 3:
Orbital radius is $r = R + h = R + 3R = 4R$.
$$v_o = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{4R}} = \frac{1}{2} \sqrt{\frac{G M}{R}} = \mathbf{\frac{v_s}{2}}$$
Correct Answer: (B)
Solution 4:
By Kepler’s Third Law:
$$\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{R}{4R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$$
$$T_1 : T_2 = \mathbf{1 : 8}$$
Correct Answer: (C)
Solution 5:
For a satellite orbiting near the surface of a planet:
$$T = \sqrt{\frac{3\pi}{G \rho}}$$
Because all factors ($3, \pi, G$) are universal constants, the minimum orbital period depends solely on the mean density $\rho$ of the planet. It is independent of the planet’s radius or mass.
Correct Answer: (C)
Solution 6:
– (A) True: $v_o = \sqrt{G M / r} \propto r^{-1/2}$.
– (B) True: $T = 2\pi \sqrt{r^3 / G M} \propto r^{3/2}$.
– (C) True: $L = m \sqrt{G M r} \propto r^{1/2}$.
– (D) True: $a_c = v_o^2 / r = (G M / r) / r = G M / r^2 \propto r^{-2}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: Friction dissipates mechanical energy: $E$ decreases.
– (B) True: $E = -G M m / (2r)$; lower energy means smaller $r$.
– (C) True: $v_o = \sqrt{G M / r}$; smaller $r$ means higher speed.
– (D) True: $K = \frac{1}{2} m v_o^2 = G M m / (2r)$; as $r$ decreases, $K$ increases.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) False: Earth’s gravitational pull is non-zero ($W = m g(h) \ne 0$).
– (B) True: Both have identical downward acceleration $a_c = g(h)$.
– (C) True: Since both accelerate identically, the normal force $N = 0$.
– (D) True: In the co-rotating frame, centrifugal force $m v^2/r$ balances gravity $m g(h)$.
Statements (B, C, D) are correct.
Correct Answer: (B, C, D)
Solution 9:
At altitude $h = R$, orbital radius is $r = 2R$.
$$v = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{2R}} = \sqrt{\frac{g_0 R^2}{2R}} = \sqrt{\frac{g_0 R}{2}}$$
Comparing with $\sqrt{\frac{g_0 R}{n}}$:
$$\mathbf{n = 2}$$
Correct Answer: 2
Solution 10:
By Kepler’s Third Law:
$$\frac{T_B}{T_A} = \left(\frac{r_B}{r_A}\right)^{3/2} = \left(\frac{9R}{R}\right)^{3/2} = (9)^{3/2} = (\sqrt{9})^3 = 3^3 = \mathbf{27}$$
Correct Answer: 27