Energy of a Satellite, Binding Energy & Orbital Transitions: Geostationary vs Polar Satellites | JEE Physics Class 11

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Concept Card: Energy of a Satellite & Geostationary vs Polar Satellites

1. Mechanics & Energy Formulation of an Orbiting Satellite:
Consider a satellite of mass $m$ revolving in a circular orbit of radius $r = R + h$ around a primary planet of mass $M$ and radius $R$ at an altitude $h$. Gravitational attraction provides the required centripetal force:
$$\frac{m v_o^2}{r} = \frac{G M m}{r^2} \implies v_o = \sqrt{\frac{G M}{r}}$$
where $v_o$ is the orbital speed.

2. Energy Components of an Orbiting Satellite

(A) Kinetic Energy ($K$):

$$K = \frac{1}{2} m v_o^2 = \frac{1}{2} m \left(\frac{G M}{r}\right) = \mathbf{\frac{G M m}{2 r} = \frac{g_0 R^2 m}{2(R + h)}}$$
Because mass and speed are positive, $K > 0$ always.

(B) Gravitational Potential Energy ($U$):

Taking the reference of zero potential energy at infinite separation ($U(\infty) = 0$):
$$U = -\int_\infty^r \vec{F}_g \cdot d\vec{r}’ = -\mathbf{\frac{G M m}{r} = -\frac{g_0 R^2 m}{R + h}}$$
Potential energy is strictly negative, reflecting the attractive nature of the gravitational field.

(C) Total Mechanical Energy ($E$):

The total mechanical energy is the sum of kinetic and potential energies:
$$E = K + U = \frac{G M m}{2 r} + \left(-\frac{G M m}{r}\right) = -\mathbf{\frac{G M m}{2 r} = -\frac{g_0 R^2 m}{2(R + h)}}$$

Fundamental Energy Relationships for Circular Orbits:
$$\mathbf{E = -K = \frac{U}{2}} \quad \Longleftrightarrow \quad \mathbf{K = |E| = -E = -\frac{U}{2}}, \quad \mathbf{U = 2 E = -2 K}$$

Physical Significance of the Negative Total Energy ($E < 0$):

  • Bound State: A negative total energy ($E < 0$) signifies that the satellite is in a gravitationally bound system (closed circular or elliptical orbit). It cannot escape to infinity on its own.
  • Unbound / Parabolic State ($E = 0$): If the satellite’s speed is increased to the escape speed $v = \sqrt{2} v_o$, total energy becomes zero, and its orbit opens into a parabola escaping to infinity.
  • Hyperbolic State ($E > 0$): If $v > \sqrt{2} v_o$, the trajectory is a hyperbola with residual kinetic energy at infinity.

3. Binding Energy & Orbital Energy Transitions

(A) Binding Energy ($BE$):

The binding energy of an orbiting satellite is the minimum energy that must be supplied to the satellite to liberate it completely from the gravitational grip of the planet to infinity ($E_f = 0$):
$$BE = E_f – E_i = 0 – E = -E = \mathbf{\frac{G M m}{2 r} = K}$$
Thus, the binding energy of a circular orbit is numerically equal to its kinetic energy.

(B) Orbital Shift / Work Required to Raise a Satellite:

To transfer a satellite from an initial circular orbit of radius $r_1$ to a higher circular orbit of radius $r_2$ ($r_2 > r_1$), work $W$ must be supplied by the rocket thrusters:
$$W = \Delta E = E_2 – E_1 = \left(-\frac{G M m}{2 r_2}\right) – \left(-\frac{G M m}{2 r_1}\right) = \mathbf{\frac{G M m}{2} \left(\frac{1}{r_1} – \frac{1}{r_2}\right)}$$
During this upward orbital transition:

  • Potential Energy Increases: $\Delta U = G M m \left(\frac{1}{r_1} – \frac{1}{r_2}\right) > 0$.
  • Kinetic Energy Decreases: $\Delta K = -\frac{G M m}{2} \left(\frac{1}{r_1} – \frac{1}{r_2}\right) < 0$.
  • Total Energy Increases: $\Delta E = \frac{1}{2} \Delta U = -\Delta K > 0$.

Notice: Although the rocket engines do positive work on the satellite, its speed actually decreases! Half the supplied energy goes into opposing kinetic energy, and the entirety converts into gravitational potential energy.

(C) Launch Energy from Earth’s Surface into Orbit $r$:

To place a satellite of mass $m$ initially at rest on the Earth’s surface ($r = R$) into an orbit of radius $r$:
$$E_{\text{surface}} = 0 + U(R) = -\frac{G M m}{R}$$
$$E_{\text{orbit}} = -\frac{G M m}{2 r}$$
$$\mathbf{E_{\text{launch}} = E_{\text{orbit}} – E_{\text{surface}} = \frac{G M m}{R} – \frac{G M m}{2 r} = G M m \left(\frac{1}{R} – \frac{1}{2 r}\right)}$$
For a low Earth orbit ($r \approx R$):
$$E_{\text{launch}} = \frac{G M m}{2 R} = \frac{1}{2} m g_0 R$$

4. Geostationary vs. Polar Satellites – Architectural Comparison

Parameter Geostationary Satellite (GEO / Communication) Polar Satellite (LEO / Remote Sensing)
Orbital Plane Equatorial plane only (Inclination $i = 0^\circ$) Passes over geographic poles (Inclination $i \approx 90^\circ$)
Sense of Rotation West to East (same as Earth’s rotation) North to South / South to North
Orbital Time Period ($T$) $T = 24\text{ hours} = 86,400\text{ s}$ (1 sidereal day) $T \approx 84\text{ to } 100\text{ minutes}$ (~14–15 orbits/day)
Orbital Altitude ($h$) $h \approx 35,800\text{ km} \approx 36,000\text{ km}$ ($r \approx 6.6 R$) $h \approx 500\text{ to } 800\text{ km}$ ($r \approx 1.1 R$)
Orbital Speed ($v_o$) $v_o \approx 3.08\text{ km/s}$ $v_o \approx 7.9\text{ to } 7.5\text{ km/s}$
Relative Motion Appears stationary to an observer on Earth Crosses different longitudes on each pass as Earth rotates
Earth Coverage Fixed view of ~one-third of Earth’s surface (cannot view polar regions) Scans the entire globe strip-by-strip over multiple days
Primary Applications Telecommunications, TV broadcasting, weather monitoring Remote sensing, meteorology, military surveillance, environmental mapping

5. Derivation of Geostationary Orbital Radius & Altitude

From Kepler’s Third Law for circular orbit of radius $r$:
$$T = 2\pi \sqrt{\frac{r^3}{G M}} \implies T^2 = \frac{4\pi^2 r^3}{G M} = \frac{4\pi^2 r^3}{g_0 R^2}$$
Solving for orbital radius $r$:
$$\mathbf{r_{\text{geo}} = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3} = \left(\frac{g_0 R^2 T^2}{4\pi^2}\right)^{1/3}}$$
Substituting Earth’s constants ($g_0 = 9.8\text{ m/s}^2, R = 6.4 \times 10^6\text{ m}, T = 86,400\text{ s}$):
$$r_{\text{geo}} = \left(\frac{9.8 \times (6.4 \times 10^6)^2 \times (86400)^2}{4 \pi^2}\right)^{1/3} \approx \mathbf{4.22 \times 10^7\text{ m} = 42,200\text{ km} \approx 6.6 R}$$
The altitude $h_{\text{geo}}$ above Earth’s surface:
$$\mathbf{h_{\text{geo}} = r_{\text{geo}} – R \approx 42,200\text{ km} – 6,400\text{ km} \approx 35,800\text{ km} \approx 36,000\text{ km}}$$

6. Critical Insights & Exam Traps in JEE:

  • Trap 1 (The Non-Equatorial Parking Fallacy): A geostationary satellite cannot be parked directly above any location away from the equator (e.g. London or New Delhi). The center of mass of the orbit must strictly coincide with the Earth’s center of mass, and the orbit must lie in the equatorial plane so that the gravitational force lies in the orbital plane without exerting out-of-plane torques.
  • Trap 2 (Work vs. Speed Paradox): When a satellite is shifted to a higher orbit, work is done on the satellite, increasing its total energy, yet its speed decreases ($v_o \propto 1/\sqrt{r}$). The potential energy increases by twice the amount of the work supplied.
  • Trap 3 (Geosynchronous vs. Geostationary):
    • Geosynchronous: Any orbit with $T = 24\text{ hours}$, regardless of inclination. If inclined, the satellite traces a figure-8 path (analemma) in the sky.
    • Geostationary: A special circular geosynchronous orbit with inclination $i = 0^\circ$ (strictly equatorial). It stays completely fixed in the sky.

Solved Examples

Example 1 (Direct Energy Calculations & Energy Ratios):
A satellite of mass $m = 1000\text{ kg}$ revolves around Earth in a circular orbit at an altitude $h = R$ above the Earth’s surface, where $R = 6.4 \times 10^6\text{ m}$ is the Earth’s radius and surface gravity is $g_0 = 9.8\text{ m/s}^2$.
(a) Find the orbital radius $r$ and the orbital speed $v_o$ of the satellite.
(b) Calculate the kinetic energy $K$, gravitational potential energy $U$, and total mechanical energy $E$ of the satellite.
(c) Determine the binding energy $BE$ required to completely free the satellite from its orbit.

Solution:
(a) Orbital Radius and Speed:
$$r = R + h = R + R = 2 R = 2(6.4 \times 10^6\text{ m}) = 1.28 \times 10^7\text{ m}$$
$$v_o = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g_0 R^2}{2 R}} = \sqrt{\frac{g_0 R}{2}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} = \sqrt{3.136 \times 10^7} = \mathbf{5600\text{ m/s} = 5.60\text{ km/s}}$$

(b) Energy Components:
Kinetic Energy:
$$K = \frac{1}{2} m v_o^2 = \frac{1}{2}(1000)(5600)^2 = \frac{1}{2}(1000)(3.136 \times 10^7) = \mathbf{1.568 \times 10^{10}\text{ J} = 15.68\text{ GJ}}$$
Potential Energy:
$$U = -\frac{G M m}{r} = -\frac{m g_0 R^2}{2 R} = -\frac{1}{2} m g_0 R = -2 K = -2(1.568 \times 10^{10}\text{ J}) = -\mathbf{3.136 \times 10^{10}\text{ J} = -31.36\text{ GJ}}$$
Total Mechanical Energy:
$$E = K + U = 1.568 \times 10^{10} – 3.136 \times 10^{10} = -\mathbf{1.568 \times 10^{10}\text{ J} = -15.68\text{ GJ}}$$

(c) Binding Energy:
$$BE = -E = K = \mathbf{1.568 \times 10^{10}\text{ J} = 15.68\text{ GJ}}$$

Example 2 (Orbital Elevation & Thruster Work Dynamics):
A satellite of mass $m$ is orbiting Earth in a stable circular orbit of radius $r_1 = 2R$. Rocket thrusters fire along the velocity vector to transfer it into an expanded circular orbit of radius $r_2 = 4R$.
(a) Determine the change in kinetic energy $\Delta K$ in terms of $m, g_0,$ and $R$.
(b) Determine the change in gravitational potential energy $\Delta U$.
(c) Find the minimum external work $W$ performed by the thrusters.
(d) Confirm the energy balance relation $\Delta E = \Delta K + \Delta U = W$.

Solution:
Recall that $G M = g_0 R^2$.
(a) Change in Kinetic Energy:
$$K_1 = \frac{G M m}{2 r_1} = \frac{g_0 R^2 m}{2(2R)} = \frac{1}{4} m g_0 R$$
$$K_2 = \frac{G M m}{2 r_2} = \frac{g_0 R^2 m}{2(4R)} = \frac{1}{8} m g_0 R$$
$$\mathbf{\Delta K = K_2 – K_1 = \frac{1}{8} m g_0 R – \frac{1}{4} m g_0 R = -\frac{1}{8} m g_0 R}$$
Kinetic energy decreases by $\frac{1}{8} m g_0 R$.

(b) Change in Potential Energy:
$$U_1 = -\frac{G M m}{r_1} = -\frac{g_0 R^2 m}{2R} = -\frac{1}{2} m g_0 R$$
$$U_2 = -\frac{G M m}{r_2} = -\frac{g_0 R^2 m}{4R} = -\frac{1}{4} m g_0 R$$
$$\mathbf{\Delta U = U_2 – U_1 = -\frac{1}{4} m g_0 R – \left(-\frac{1}{2} m g_0 R\right) = +\frac{1}{4} m g_0 R}$$
Potential energy increases by $\frac{1}{4} m g_0 R$.

(c) Minimum Work Performed by Thrusters:
$$W = \Delta E = E_2 – E_1 = \left(-\frac{G M m}{2 r_2}\right) – \left(-\frac{G M m}{2 r_1}\right) = \frac{G M m}{2} \left(\frac{1}{2R} – \frac{1}{4R}\right) = \frac{g_0 R^2 m}{2} \left(\frac{1}{4R}\right) = \mathbf{+\frac{1}{8} m g_0 R}$$

(d) Energy Balance Verification:
$$\Delta K + \Delta U = -\frac{1}{8} m g_0 R + \frac{1}{4} m g_0 R = +\frac{1}{8} m g_0 R = W = \Delta E$$
The energy accounting is precisely verified.

Example 3 (Geostationary Satellite Parameter Scaling & Planetary Day Alteration):
Currently, a geostationary satellite has an orbital radius $r_0 \approx 42,000\text{ km}$ corresponding to an Earth rotational period $T_0 = 24\text{ hours}$.
(a) Derive the relationship showing how the geostationary orbital radius scales with the planetary rotational period $T$.
(b) If due to internal core redistribution, the Earth’s rotation rate accelerates such that one complete day lasts only $T’ = 8\text{ hours}$, calculate the new geostationary orbital radius $r’$ and its altitude $h’$ above the Earth’s surface ($R = 6400\text{ km}$).

Solution:
(a) Scaling Law:
From Kepler’s Third Law for circular motion:
$$T = 2\pi \sqrt{\frac{r^3}{G M}} \implies T^2 \propto r^3 \implies \mathbf{r \propto T^{2/3}}$$

(b) New Orbital Radius for $T’ = 8\text{ hours}$:
$$\frac{r’}{r_0} = \left(\frac{T’}{T_0}\right)^{2/3} = \left(\frac{8}{24}\right)^{2/3} = \left(\frac{1}{3}\right)^{2/3} = 3^{-2/3} = \frac{1}{(9)^{1/3}} \approx \frac{1}{2.080} \approx 0.4807$$
$$r’ = (0.4807)(42,000\text{ km}) \approx \mathbf{20,190\text{ km}}$$
The new altitude above the surface:
$$h’ = r’ – R = 20,190\text{ km} – 6,400\text{ km} = \mathbf{13,790\text{ km}}$$

Example 4 (Polar Satellite Ground Track & Equatorial Strip Scanning Dynamics):
A remote-sensing polar satellite orbits in a circular sun-synchronous orbit at an altitude $h = 600\text{ km}$ above the Earth’s surface ($R = 6400\text{ km}, g_0 = 9.8\text{ m/s}^2$).
(a) Calculate the orbital period $T$ of the satellite in minutes.
(b) Calculate the angle through which the Earth rotates about its axis during one complete revolution of the satellite.
(c) Determine the longitudinal ground-track distance (strip width) traversed at the equator between two consecutive passes of the satellite.

Solution:
(a) Orbital Period:
Orbital radius: $r = R + h = 6400\text{ km} + 600\text{ km} = 7000\text{ km} = 7.0 \times 10^6\text{ m}$.
$$T = 2\pi \sqrt{\frac{r^3}{G M}} = 2\pi \sqrt{\frac{r^3}{g_0 R^2}} = \frac{2\pi}{R} \sqrt{\frac{r^3}{g_0}} = \frac{2\pi}{6.4 \times 10^6} \sqrt{\frac{(7.0 \times 10^6)^3}{9.8}}$$
$$(7.0 \times 10^6)^3 / 9.8 = 3.43 \times 10^{20} / 9.8 = 3.50 \times 10^{19}$$
$$\sqrt{3.50 \times 10^{19}} = 5.916 \times 10^9$$
$$T = \frac{2\pi \times 5.916 \times 10^9}{6.4 \times 10^6} = \frac{3.717 \times 10^{10}}{6.4 \times 10^6} \approx 5808\text{ seconds}$$
In minutes:
$$\mathbf{T = \frac{5808}{60} \approx 96.8\text{ minutes} \approx 1\text{ hour } 36.8\text{ min}}$$

(b) Earth Rotation Angle per Orbit:
Earth rotates $360^\circ$ in $T_E = 24\text{ hours} = 86,400\text{ s}$.
$$\Delta \theta = \left(\frac{T}{T_E}\right) \times 360^\circ = \left(\frac{5808\text{ s}}{86400\text{ s}}\right) \times 360^\circ = \mathbf{24.2^\circ}$$

(c) Equatorial Strip Width:
The linear distance along the equator corresponding to angle $\Delta \theta$:
$$D_{\text{eq}} = R \cdot \Delta \theta_{\text{rad}} = R \left(24.2 \times \frac{\pi}{180}\right) = (6400\text{ km})(0.4224) \approx \mathbf{2703\text{ km}}$$
Each successive north-to-south pass shifts westward by $2703\text{ km}$ at the equator.

Practice Worksheet: Energy of a Satellite & Geostationary vs Polar Satellites

Section A: Single Correct Questions (Q1 to Q5)

Q1. A satellite of mass $m$ is revolving in a circular orbit of radius $r$ around Earth. If $K, U,$ and $E$ represent its kinetic energy, gravitational potential energy, and total mechanical energy respectively (with $U = 0$ at infinity), which of the following relations is correct?
(A) $E = -K = \frac{U}{2}$
(B) $E = K = -\frac{U}{2}$
(C) $E = -K = 2 U$
(D) $E = -2 K = U$

Q2. A satellite of mass $m$ is orbiting Earth in a circular orbit of radius $r_1 = 2R$. The minimum work required to transfer this satellite into an expanded circular orbit of radius $r_2 = 3R$ is (where $g_0$ is surface gravity):
(A) $\frac{1}{6} m g_0 R$
(B) $\frac{1}{12} m g_0 R$
(C) $\frac{1}{3} m g_0 R$
(D) $\frac{1}{24} m g_0 R$

Q3. The minimum energy required to launch a satellite of mass $m$ from the surface of Earth ($r = R$) into a stable circular orbit of radius $r = 2R$ is:
(A) $\frac{1}{4} m g_0 R$
(B) $\frac{1}{2} m g_0 R$
(C) $\frac{3}{4} m g_0 R$
(D) $m g_0 R$

Q4. A geostationary satellite orbits at a distance $r$ from the center of Earth. If the angular velocity of the Earth’s rotation were doubled, the new orbital radius of a geostationary satellite would be:
(A) $\frac{r}{2}$
(B) $\frac{r}{4^{1/3}}$
(C) $\frac{r}{\sqrt{2}}$
(D) $2^{2/3} r$

Q5. Why can a geostationary satellite NOT be parked directly over an observer located at a latitude of $45^\circ\text{ N}$?
(A) The gravitational attraction of Earth at higher latitudes is insufficient to supply the centripetal acceleration.
(B) The center of every orbital plane must coincide with the center of the Earth, so an orbit fixed above a non-equatorial latitude is physically impossible.
(C) Atmospheric drag in non-equatorial upper regions quickly decays the satellite orbit.
(D) The Coriolis force in the northern hemisphere pushes the satellite toward the equator.

Section B: One or More than One Correct Questions (Q6 to Q8)

Q6. For a satellite describing a circular orbit of radius $r$ around Earth, which of the following statements is/are correct?
(A) The kinetic energy is positive, potential energy is negative, and total mechanical energy is negative.
(B) As the orbital radius $r$ increases, the total mechanical energy increases while the orbital speed decreases.
(C) The binding energy required to liberate the satellite to infinity is equal to its kinetic energy.
(D) If an orbiting satellite is supplied with additional energy equal to its kinetic energy, its total energy becomes zero and it escapes along a parabolic trajectory.

Q7. Which of the following statements regarding geostationary and polar satellites is/are correct?
(A) A geostationary satellite has an orbital period of 24 hours and revolves from West to East strictly in the equatorial plane.
(B) Polar satellites orbit at low altitudes ($500 – 800\text{ km}$) with periods of approximately $84 – 100\text{ minutes}$.
(C) Polar satellites can scan the entire surface of the Earth strip by strip because the Earth rotates beneath their north-south orbital plane.
(D) A satellite placed in an orbit inclined at $45^\circ$ to the equator with a 24-hour orbital period remains completely stationary over a fixed point on Earth.

Q8. 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 gradually decreases.
(C) The linear orbital speed $v_o$ of the satellite increases.
(D) The rate of loss of potential energy is twice the rate of gain of kinetic energy.

Section C: Numerical Value Type Questions (Q9 to Q10)

Q9. A satellite is transferred from a circular orbit of radius $r$ to a circular orbit of radius $2r$. The ratio of the work done by the thrusters to the initial kinetic energy of the satellite in the original orbit is equal to $\frac{1}{n}$. Find the integer value of $n$.

Q10. A geostationary satellite has an orbital radius $r_{\text{geo}}$ when the Earth’s rotational period is $24\text{ hours}$. If Earth were to spin faster such that its rotational period became $3\text{ hours}$, the new geostationary orbital radius would become $r’ = \frac{r_{\text{geo}}}{k}$. Determine the integer value of $k$.

Comprehensive Solutions and Answer Key

Answer Key

Question Correct Answer Core Physics Concept
Q1 (A) Fundamental Energy Relationship: $E = -K = U/2$
Q2 (B) Orbital Elevation Work: $W = \Delta E = \frac{mg_0 R}{12}$
Q3 (C) Surface Launch Energy: $E_{\text{launch}} = E_{\text{orbit}} – E_{\text{surf}} = \frac{3}{4} mg_0 R$
Q4 (B) Angular Velocity Scaling: $r \propto \omega^{-2/3} = r / 4^{1/3}$
Q5 (B) Orbital Mechanics: Orbital Center Must Coincide with Earth Center
Q6 (A, B, C, D) Complete Energy and Escape Dynamics of Circular Orbit
Q7 (A, B, C) Geostationary vs Polar Satellite Characteristics
Q8 (A, B, C, D) Atmospheric Drag Paradox: Kinetic Energy Increases as $r$ Decays
Q9 2 Work-to-Kinetic Ratio: $W / K_1 = (GMm/4r) / (GMm/2r) = 1/2 \implies n = 2$
Q10 4 Period Reduction Scaling: $(3/24)^{2/3} = (1/8)^{2/3} = 1/4 \implies k = 4$

Detailed Explanations

Solution 1:
For a satellite of mass $m$ in a circular orbit of radius $r$:
– Kinetic energy: $K = \frac{G M m}{2 r} > 0$
– Potential energy: $U = -\frac{G M m}{r} < 0$ - Total energy: $E = K + U = -\frac{G M m}{2 r} < 0$ Comparing these expressions: $$E = -K = \frac{U}{2}$$ Correct Answer: (A)

Solution 2:
The work required to transfer the satellite from $r_1 = 2R$ to $r_2 = 3R$ is equal to the change in total mechanical energy:
$$W = \Delta E = E_2 – E_1 = -\frac{G M m}{2(3R)} – \left(-\frac{G M m}{2(2R)}\right) = \frac{G M m}{2 R} \left(\frac{1}{2} – \frac{1}{3}\right) = \frac{G M m}{2 R} \left(\frac{1}{6}\right) = \frac{G M m}{12 R}$$
Since $G M = g_0 R^2$:
$$W = \frac{m g_0 R^2}{12 R} = \mathbf{\frac{1}{12} m g_0 R}$$
Correct Answer: (B)

Solution 3:
Initial mechanical energy at the Earth’s surface ($r = R$) for a resting satellite:
$$E_i = 0 + U(R) = -\frac{G M m}{R}$$
Final total mechanical energy in circular orbit of radius $r = 2R$:
$$E_f = -\frac{G M m}{2(2R)} = -\frac{G M m}{4 R}$$
The minimum launch energy required:
$$E_{\text{launch}} = E_f – E_i = -\frac{G M m}{4 R} – \left(-\frac{G M m}{R}\right) = \frac{3 G M m}{4 R} = \mathbf{\frac{3}{4} m g_0 R}$$
Correct Answer: (C)

Solution 4:
For a geostationary satellite, the orbital period equals the rotational period of Earth:
$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{r^3}{G M}} \implies r^3 = \frac{G M}{\omega^2} \implies r \propto \omega^{-2/3}$$
If $\omega$ is doubled ($\omega’ = 2\omega$):
$$r’ = r \left(\frac{\omega}{2\omega}\right)^{2/3} = r \left(\frac{1}{2}\right)^{2/3} = \frac{r}{2^{2/3}} = \mathbf{\frac{r}{4^{1/3}}}$$
Correct Answer: (B)

Solution 5:
Newton’s law of universal gravitation dictates that the central gravitational force acts directly toward the center of the Earth. Therefore, the plane of every planetary orbit must pass through the center of the Earth. A plane passing through a parallel of latitude (like $45^\circ\text{ N}$) does not contain the Earth’s center; the gravitational force would have a component perpendicular to that plane pulling the satellite toward the equator. Hence, a stationary orbit over a non-equatorial latitude is physically impossible.
Correct Answer: (B)

Solution 6:
– (A) True: $K = +G M m / (2r) > 0$, $U = -G M m / r < 0$, $E = -G M m / (2r) < 0$. - (B) True: As $r \to \infty$, $E$ approaches 0 from negative values (increases), while $v_o = \sqrt{G M / r}$ decreases. - (C) True: Binding energy $BE = -E = K$. - (D) True: Adding energy $\Delta E = K$ makes $E_{\text{total}} = -K + K = 0$, which corresponds to the parabolic escape threshold ($v = \sqrt{2} v_o$).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)

Solution 7:
– (A) True: Geostationary orbit must be equatorial with $T = 24\text{ h}$ revolving west to east.
– (B) True: Polar satellites are low Earth orbiters ($500 – 800\text{ km}$) with periods of $84 – 100\text{ minutes}$.
– (C) True: Earth’s rotation under the north-south polar orbit facilitates global strip-by-strip coverage.
– (D) False: An inclined 24-hour orbit is geosynchronous, not geostationary; the satellite moves north and south in the sky tracing a figure-8 analemma.
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)

Solution 8:
– (A) True: Drag dissipates mechanical energy ($W_{\text{drag}} < 0 \implies \Delta E < 0$). - (B) True: Since $E = -G M m / (2r)$, a decrease in $E$ means $r$ must decrease. - (C) True: Since $v_o = \sqrt{G M / r}$, as $r$ decreases, $v_o$ increases. - (D) True: $dU/dr = G M m / r^2$, $dK/dr = -G M m / (2r^2) = -\frac{1}{2} dU/dr$. Therefore, $|\Delta U| = 2 \Delta K$. Potential energy decreases twice as fast as kinetic energy increases, with the difference dissipated as frictional heat.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)

Solution 9:
Initial kinetic energy at radius $r$:
$$K_1 = \frac{G M m}{2 r}$$
Work done to transfer from $r$ to $2r$:
$$W = E_2 – E_1 = -\frac{G M m}{2(2r)} – \left(-\frac{G M m}{2 r}\right) = \frac{G M m}{2 r} \left(1 – \frac{1}{2}\right) = \frac{G M m}{4 r}$$
Taking the ratio:
$$\frac{W}{K_1} = \frac{G M m / (4 r)}{G M m / (2 r)} = \frac{1}{2}$$
Comparing with $\frac{1}{n}$:
$$\mathbf{n = 2}$$
Correct Answer: 2

Solution 10:
Using Kepler’s Third Law $r \propto T^{2/3}$:
$$\frac{r’}{r_{\text{geo}}} = \left(\frac{T’}{T}\right)^{2/3} = \left(\frac{3\text{ h}}{24\text{ h}}\right)^{2/3} = \left(\frac{1}{8}\right)^{2/3} = \left(\frac{1}{2^3}\right)^{2/3} = \frac{1}{2^2} = \frac{1}{4}$$
$$r’ = \frac{r_{\text{geo}}}{4} \implies \mathbf{k = 4}$$
Correct Answer: 4

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