Concept Card: Gravitational Potential and Escape Velocity – Derivations & Dynamics
1. Fundamental Concept of Gravitational Potential ($V$):
While gravitational force $\vec{F}_g$ and field $\vec{E}_g$ are vector quantities describing gravitational attraction per unit mass, gravitational potential ($V$) is their scalar counterpart. It defines the energy state of space surrounding massive bodies.
Rigorous Physical Definition:
The gravitational potential at any point in a gravitational field is defined as the work done by an external agent per unit mass in bringing a small test mass from infinity to that point slowly (without imparting any kinetic energy):
$$V(\vec{r}) = \frac{W_{\text{ext}}}{m} = -\int_\infty^{\vec{r}} \vec{E}_g \cdot d\vec{r}’ = \frac{U(\vec{r})}{m}$$
- SI Unit: $\text{J/kg}$ or $\text{m}^2/\text{s}^2$.
- Dimensional Formula: $[V] = \frac{[W]}{[m]} = \frac{[M L^2 T^{-2}]}{[M]} = \mathbf{[M^0 L^2 T^{-2}]}$.
- Scalar Nature: As a scalar, potentials due to multiple source masses add algebraically: $V_{\text{net}} = \sum_i V_i$.
- Relationship to Gravitational Field:
$$\vec{E}_g = -\vec{\nabla} V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right) \implies E_r = -\frac{dV}{dr}$$
The gravitational field always points in the direction of the steepest decrease of gravitational potential.
2. Gravitational Potential Profiles of Continuous Spherically Symmetric Bodies
(A) Uniform Thin Spherical Shell of Mass $M$ and Radius $R$:
- Exterior Points ($r \ge R$):
$$\mathbf{V(r) = -\frac{G M}{r}}$$ - Surface ($r = R$):
$$V_s = -\frac{G M}{R}$$ - Interior Points ($r < R$): Since the gravitational field inside a uniform spherical shell is identically zero ($\vec{E}_g = \vec{0}$), the potential gradient vanishes ($\frac{dV}{dr} = 0$). Hence, the interior is an equipotential volume equal to the surface potential:
$$\mathbf{V(r) = -\frac{G M}{R} = \text{constant} \quad (\text{for all } r \le R)}$$
(B) Uniform Solid Sphere of Mass $M$ and Radius $R$:
- Exterior Points ($r \ge R$):
$$\mathbf{V_{\text{out}}(r) = -\frac{G M}{r}}$$ - Surface ($r = R$):
$$V_s = -\frac{G M}{R}$$ - Interior Points ($r \le R$):
Integrating from the surface inward where $E_{\text{in}}(r’) = -\frac{G M r’}{R^3}$:
$$V_{\text{in}}(r) – V(R) = -\int_R^r E_{\text{in}}(r’) dr’ = -\int_R^r \left(-\frac{G M r’}{R^3}\right) dr’ = \frac{G M}{R^3} \left[ \frac{r^2 – R^2}{2} \right]$$
$$V_{\text{in}}(r) = -\frac{G M}{R} + \frac{G M}{2 R^3}(r^2 – R^2) = -\frac{G M}{2 R^3} (2 R^2 + R^2 – r^2)$$Internal Potential of a Uniform Solid Sphere ($r \le R$):
$$\mathbf{V_{\text{in}}(r) = -\frac{G M}{2 R^3} \left(3 R^2 – r^2\right)}$$ - At the Center ($r = 0$):
$$\mathbf{V_{\text{center}} = -\frac{3 G M}{2 R} = 1.5 V_{\text{surface}}}$$
The gravitational potential at the center is $1.5$ times deeper than at the surface.
3. Rigorous Derivation of Escape Velocity ($v_e$)
Escape velocity is the minimum projection speed required for a particle to escape permanently from the gravitational field of a celestial body to infinite separation without any subsequent thrust.
Derivation via the Work-Energy Theorem:
Consider a particle of mass $m$ projected with initial speed $v_e$ from a point at distance $r$ from the center of a celestial body of mass $M$.
– Initial Mechanical Energy at Distance $r$:
$$E_i = K_i + U_i = \frac{1}{2} m v_e^2 + m V(r)$$
– Final Mechanical Energy at Infinity ($r \to \infty$):
To achieve escape with the absolute minimum speed, the particle arrives at infinity with zero residual kinetic energy ($v_\infty = 0$), where potential energy is zero ($U(\infty) = 0$):
$$E_f = K_f + U_f = 0 + 0 = 0$$
By Conservation of Mechanical Energy ($E_i = E_f$):
$$\frac{1}{2} m v_e^2 + m V(r) = 0 \implies \mathbf{v_e = \sqrt{-2 V(r)}}$$
(A) Escape Velocity from the Earth’s Surface ($r = R$):
At the surface, $V_s = -\frac{G M}{R}$. Substituting $V_s$:
$$v_e = \sqrt{-2\left(-\frac{G M}{R}\right)} = \mathbf{\sqrt{\frac{2 G M}{R}}}$$
Since $g_0 = \frac{G M}{R^2} \implies G M = g_0 R^2$:
$$\mathbf{v_e = \sqrt{2 g_0 R} = \sqrt{2 \left(\frac{4}{3}\pi G \rho R\right) R} = R \sqrt{\frac{8}{3}\pi G \rho}}$$
Numerical Value for Earth: With $R = 6.4 \times 10^6\text{ m}$ and $g_0 = 9.8\text{ m/s}^2$:
$$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.2544 \times 10^8} \approx \mathbf{11.2\text{ km/s}}$$
(B) Fundamental Invariance Properties of Escape Velocity:
- Independence of Projectile Mass ($m$): An electron, a bowling ball, and a Saturn V rocket have the exact same escape velocity from Earth ($11.2\text{ km/s}$).
- Independence of Projection Angle ($\theta$): Because escape velocity is derived purely from scalar energy conservation, it is independent of the direction or launch angle $\theta$, provided the trajectory does not physically collide with the planetary surface.
4. Escape Velocity in Diverse Physical Contexts
- From an Altitude $h$ Above Earth’s Surface ($r = R + h$):
$$\mathbf{v_e(h) = \sqrt{\frac{2 G M}{R + h}} = \sqrt{\frac{2 g_0 R^2}{R + h}} = v_{e, 0} \sqrt{\frac{R}{R + h}}}$$ - From the Center of the Earth ($r = 0$):
$$V_{\text{center}} = -\frac{3 G M}{2 R} \implies v_{e, \text{center}} = \sqrt{-2\left(-\frac{3 G M}{2 R}\right)} = \mathbf{\sqrt{\frac{3 G M}{R}} = \sqrt{3 g_0 R} = \sqrt{1.5} v_{e, 0} \approx 13.72\text{ km/s}}$$ - Residual Velocity at Infinity ($v_\infty$) when Projected with Speed $v > v_e$:
$$\frac{1}{2} m v^2 – \frac{G M m}{R} = \frac{1}{2} m v_\infty^2 + 0 \implies v^2 – v_e^2 = v_\infty^2 \implies \mathbf{v_\infty = \sqrt{v^2 – v_e^2}}$$
5. Critical Insights & Exam Traps in JEE:
- Trap 1 (Escape Speed vs. Orbital Speed):
– Orbital speed in low Earth orbit: $v_{\text{orb}} = \sqrt{\frac{G M}{R}} = \sqrt{g_0 R} \approx 7.92\text{ km/s}$.
– Escape speed from surface: $v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g_0 R} \approx 11.2\text{ km/s}$.
$$\mathbf{v_e = \sqrt{2} v_{\text{orb}} \approx 1.414 v_{\text{orb}}}$$
Increasing a circular satellite’s speed by a mere $41.4\%$ ($\Delta v = (\sqrt{2} – 1) v_{\text{orb}}$) shifts its orbit from circular to parabolic escape! - Trap 2 (Launch Angle Fallacy): A projectile fired at $30^\circ, 45^\circ,$ or $90^\circ$ to the horizontal requires the exact same speed $11.2\text{ km/s}$ to escape to infinity.
- Trap 3 (Gravitational Potential at Center): Potential at the center is not zero. Although $\vec{E}_g = \vec{0}$ at the center, potential is at its deepest negative value: $V_c = -1.5 \frac{G M}{R}$.
Solved Examples
Example 1 (Direct Planetary Comparison & Scaling Laws):
A hypothetical exoplanet has eight times the mass of the Earth ($M_p = 8 M_e$) and twice the radius of the Earth ($R_p = 2 R_e$). Given that the escape speed from Earth is $v_{ee} = 11.2\text{ km/s}$ and surface gravity is $g_e = 9.8\text{ m/s}^2$:
(a) Find the ratio of the surface gravitational potential of the exoplanet to that of Earth ($V_p / V_e$).
(b) Find the ratio of the surface gravitational acceleration of the exoplanet to that of Earth ($g_p / g_e$).
(c) Calculate the escape velocity $v_{ep}$ from the surface of the exoplanet.
Solution:
(a) Gravitational Potential Ratio:
$$V = -\frac{G M}{R} \implies \frac{V_p}{V_e} = \left(\frac{M_p}{M_e}\right) \left(\frac{R_e}{R_p}\right) = (8) \left(\frac{1}{2}\right) = \mathbf{4}$$
The potential well at the exoplanet’s surface is 4 times deeper than Earth’s.
(b) Surface Gravity Ratio:
$$g = \frac{G M}{R^2} \implies \frac{g_p}{g_e} = \left(\frac{M_p}{M_e}\right) \left(\frac{R_e}{R_p}\right)^2 = (8) \left(\frac{1}{2}\right)^2 = \frac{8}{4} = \mathbf{2}$$
Surface gravity on the exoplanet is $2 g_e = 19.6\text{ m/s}^2$.
(c) Escape Velocity:
$$v_e = \sqrt{\frac{2 G M}{R}} \implies \frac{v_{ep}}{v_{ee}} = \sqrt{\frac{M_p / M_e}{R_p / R_e}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$
$$v_{ep} = 2 v_{ee} = 2(11.2\text{ km/s}) = \mathbf{22.4\text{ km/s}}$$
Example 2 (Hyperbolic Escape & Interplanetary Residual Velocity):
A deep-space probe is launched from the Earth’s surface with an initial velocity $v_0 = 2 v_e$, where $v_e = 11.2\text{ km/s}$ is the surface escape speed.
(a) Determine the residual asymptotic speed $v_\infty$ of the probe when it escapes entirely from the Earth’s gravitational field.
(b) Calculate the speed $v(r)$ of the probe when it reaches a radial distance $r = 4R$ from the center of the Earth.
Solution:
(a) Asymptotic Speed at Infinity:
By conservation of mechanical energy:
$$\frac{1}{2} m v_0^2 – \frac{G M m}{R} = \frac{1}{2} m v_\infty^2 + 0$$
Substitute $\frac{G M m}{R} = \frac{1}{2} m v_e^2$:
$$\frac{1}{2} m v_0^2 – \frac{1}{2} m v_e^2 = \frac{1}{2} m v_\infty^2 \implies v_\infty^2 = v_0^2 – v_e^2$$
Given $v_0 = 2 v_e$:
$$v_\infty^2 = (2 v_e)^2 – v_e^2 = 4 v_e^2 – v_e^2 = 3 v_e^2$$
$$\mathbf{v_\infty = \sqrt{3} v_e} = \sqrt{3}(11.2\text{ km/s}) \approx 1.732 \times 11.2 = \mathbf{19.40\text{ km/s}}$$
(b) Speed at Distance $r = 4R$:
$$\frac{1}{2} m v(4R)^2 – \frac{G M m}{4R} = \frac{1}{2} m v_0^2 – \frac{G M m}{R}$$
$$\frac{1}{2} v(4R)^2 = \frac{1}{2} v_0^2 – \frac{G M}{R} + \frac{G M}{4R} = \frac{1}{2}(4 v_e^2) – \frac{1}{2} v_e^2 + \frac{1}{8} v_e^2$$
$$v(4R)^2 = 4 v_e^2 – v_e^2 + \frac{1}{4} v_e^2 = 3.25 v_e^2 = \frac{13}{4} v_e^2$$
$$\mathbf{v(4R) = \frac{\sqrt{13}}{2} v_e} = \frac{3.6055}{2}(11.2\text{ km/s}) \approx \mathbf{20.19\text{ km/s}}$$
Example 3 (Multi-Concept Linkage – Escape Dynamics from the Center of Earth):
A particle of mass $m$ is located at the center of the Earth ($r = 0$).
(a) Find the minimum speed $u_1$ required to propel the particle through a frictionless tunnel to reach the Earth’s surface ($r = R$).
(b) Find the minimum speed $u_2$ required to propel the particle from the center so that it escapes permanently to infinity.
(c) If a particle falls from rest at infinity toward Earth along the tunnel axis, find its speed as it crosses the Earth’s center.
Solution:
(a) Speed to Reach Surface from Center:
By conservation of mechanical energy between $r = 0$ and $r = R$:
$$\frac{1}{2} m u_1^2 + m V_{\text{center}} = 0 + m V_{\text{surface}}$$
$$\frac{1}{2} u_1^2 + \left(-\frac{3 G M}{2 R}\right) = -\frac{G M}{R}$$
$$\frac{1}{2} u_1^2 = \frac{3 G M}{2 R} – \frac{G M}{R} = \frac{G M}{2 R} \implies u_1^2 = \frac{G M}{R} = g_0 R$$
$$\mathbf{u_1 = \sqrt{g_0 R}} = \frac{v_e}{\sqrt{2}} \approx \mathbf{7.92\text{ km/s}}$$
(b) Escape Speed from the Center to Infinity:
Between $r = 0$ and $r = \infty$:
$$\frac{1}{2} m u_2^2 + m V_{\text{center}} = 0 + 0$$
$$\frac{1}{2} u_2^2 = -V_{\text{center}} = \frac{3 G M}{2 R} \implies u_2^2 = \frac{3 G M}{R} = 1.5 \left(\frac{2 G M}{R}\right) = 1.5 v_e^2$$
$$\mathbf{u_2 = \sqrt{3 g_0 R} = \sqrt{1.5} v_e} = \sqrt{1.5}(11.2\text{ km/s}) \approx \mathbf{13.72\text{ km/s}}$$
(c) Free-Fall Infall from Infinity:
$$\frac{1}{2} m v^2 + m V_{\text{center}} = 0 \implies v = \sqrt{-2 V_{\text{center}}} = \mathbf{\sqrt{3 g_0 R} \approx 13.72\text{ km/s}}$$
(The inbound infall speed at the center equals the outbound launch escape speed from the center!).
Example 4 (Continuous Mass Integration – Potential & Escape Speed from a Thin Ring):
A uniform thin circular wire ring of mass $M$ and radius $R$ rests in a fixed plane.
(a) Determine the gravitational potential $V(x)$ at an axial point at distance $x$ from the center of the ring.
(b) Calculate the minimum escape speed $v_{e, \text{center}}$ required to shoot a particle along the ring axis to infinity from the center of the ring ($x = 0$).
(c) Calculate the escape speed $v_e(x)$ from an axial distance $x = R\sqrt{3}$.
Solution:
(a) Gravitational Potential on the Ring Axis:
Every infinitesimal element $dM$ of the ring is at the exact same distance $r = \sqrt{R^2 + x^2}$ from the axial point.
$$V(x) = \int -\frac{G dM}{\sqrt{R^2 + x^2}} = -\frac{G}{\sqrt{R^2 + x^2}} \int dM = -\mathbf{\frac{G M}{\sqrt{R^2 + x^2}}}$$
(b) Escape Speed from the Center ($x = 0$):
At $x = 0$, $V(0) = -\frac{G M}{R}$.
Conservation of energy to infinity:
$$\frac{1}{2} m v_{e, \text{center}}^2 + m V(0) = 0 \implies v_{e, \text{center}} = \sqrt{-2 V(0)} = \mathbf{\sqrt{\frac{2 G M}{R}}}$$
(c) Escape Speed from $x = R\sqrt{3}$:
Distance to the ring perimeter:
$$r = \sqrt{R^2 + (R\sqrt{3})^2} = \sqrt{R^2 + 3 R^2} = \sqrt{4 R^2} = 2 R$$
Potential at this point:
$$V(R\sqrt{3}) = -\frac{G M}{2 R}$$
Escape velocity:
$$\frac{1}{2} m v_e^2 + m V(R\sqrt{3}) = 0 \implies v_e = \sqrt{-2 V(R\sqrt{3})} = \sqrt{2 \left(\frac{G M}{2 R}\right)} = \mathbf{\sqrt{\frac{G M}{R}}}$$
Practice Worksheet: Gravitational Potential & Escape Velocity
Section A: Single Correct Questions (Q1 to Q5)
Q1. If $V_s$ denotes the gravitational potential at the surface of a uniform solid sphere of radius $R$, the gravitational potential at its center $V_c$ is equal to:
(A) Zero
(B) $V_s$
(C) $\frac{3}{2} V_s$
(D) $2 V_s$
Q2. A projectile is launched vertically upwards from the surface of Earth with a speed equal to half the escape velocity ($v = v_e / 2$). The maximum height $h$ attained above the Earth’s surface is:
(A) $\frac{R}{4}$
(B) $\frac{R}{3}$
(C) $\frac{R}{2}$
(D) $R$
Q3. The escape velocity from the surface of Earth for a body of mass $m$ is $11.2\text{ km/s}$. The escape velocity for another body of mass $4m$ launched from the same location is:
(A) $5.6\text{ km/s}$
(B) $11.2\text{ km/s}$
(C) $22.4\text{ km/s}$
(D) $44.8\text{ km/s}$
Q4. The escape velocity of a body from an altitude $h = 3R$ above the surface of Earth (where $R$ is Earth’s radius) in terms of the surface escape velocity $v_e$ is:
(A) $\frac{v_e}{2}$
(B) $\frac{v_e}{\sqrt{2}}$
(C) $\frac{v_e}{3}$
(D) $\frac{v_e}{4}$
Q5. A body is projected from Earth’s surface at an angle of $30^\circ$ to the vertical. The minimum projection speed required for it to escape Earth’s gravity is:
(A) $11.2\text{ km/s}$
(B) $11.2 \cos 30^\circ\text{ km/s}$
(C) $11.2 \sin 30^\circ\text{ km/s}$
(D) $11.2 / \cos 30^\circ\text{ km/s}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding the gravitational potential of a uniform solid sphere of mass $M$ and radius $R$ is/are correct?
(A) The gravitational potential is continuous everywhere across $r = R$.
(B) The gravitational field $\vec{E}_g = -\frac{dV}{dr}\hat{r}$ is continuous everywhere across $r = R$.
(C) The potential at the center $r = 0$ is minimum in value (deepest negative potential).
(D) The potential is constant throughout the interior of the sphere.
Q7. A rocket is launched from Earth with speed $v > v_e$. Which of the following statements is/are correct?
(A) Its trajectory relative to Earth is a hyperbola.
(B) Its speed at infinite distance from Earth is $v_\infty = \sqrt{v^2 – v_e^2}$.
(C) Total mechanical energy of the rocket is positive.
(D) The rocket will continue to accelerate as it recedes from Earth.
Q8. In relativistic astrophysics, the Schwarzschild radius $R_s$ of a black hole of mass $M$ is estimated by setting the Newtonian escape velocity equal to the speed of light $c$. Which of the following statements is/are correct?
(A) $R_s = \frac{2 G M}{c^2}$.
(B) The Schwarzschild radius of Earth ($M \approx 6 \times 10^{24}\text{ kg}$) is approximately $9\text{ mm}$.
(C) The average mass density within the Schwarzschild radius is inversely proportional to $M^2$.
(D) Nothing, not even light, can escape from within $r < R_s$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A body is projected vertically upwards from the surface of Earth with a speed equal to three times the escape velocity ($v = 3 v_e$). Its residual speed when it reaches infinite separation from Earth is $v_\infty = \sqrt{n} v_e$. Find the integer value of $n$.
Q10. The ratio of the escape velocity of a particle from the center of the Earth through a tunnel to the escape velocity from the surface of the Earth is equal to $\sqrt{\frac{k}{2}}$. Determine the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (C) | Central Potential of Solid Sphere: $V_c = \frac{3}{2} V_s$ |
| Q2 | (B) | Sub-Escape Launch Altitude: $\frac{1}{4} = \frac{h/R}{1 + h/R} \implies h = R/3$ |
| Q3 | (B) | Mass Independence of Escape Velocity ($v_e = \sqrt{2 g_0 R}$) |
| Q4 | (A) | Altitude Scaling: $v_e(3R) = \sqrt{\frac{2GM}{4R}} = \frac{1}{2} v_e$ |
| Q5 | (A) | Angle Independence of Escape Energy Threshold |
| Q6 | (A, B, C) | Continuity & Shape of Solid Sphere Potential Function |
| Q7 | (A, B, C) | Hyperbolic Orbit Dynamics & Residual Kinetic Energy |
| Q8 | (A, B, C, D) | Schwarzschild Black Hole Radius Derivation ($R_s = 2GM/c^2$) |
| Q9 | 8 | Asymptotic Velocity Formula: $v_\infty = \sqrt{3^2 – 1} v_e = \sqrt{8} v_e$ |
| Q10 | 3 | Center vs Surface Escape Ratio: $\sqrt{3 g_0 R} / \sqrt{2 g_0 R} = \sqrt{3/2}$ |
Detailed Explanations
Solution 1:
For a uniform solid sphere of mass $M$ and radius $R$:
– At surface: $V_s = -\frac{G M}{R}$.
– At center: $V_c = -\frac{3 G M}{2 R}$.
$$\frac{V_c}{V_s} = \frac{-1.5 G M / R}{-G M / R} = \frac{3}{2} \implies \mathbf{V_c = \frac{3}{2} V_s}$$
Correct Answer: (C)
Solution 2:
Initial kinetic energy at surface:
$$K_i = \frac{1}{2} m \left(\frac{v_e}{2}\right)^2 = \frac{1}{2} m \frac{v_e^2}{4} = \frac{1}{8} m (2 g_0 R) = \frac{1}{4} m g_0 R$$
The gain in potential energy upon ascending to altitude $h$ is:
$$\Delta U = \frac{m g_0 h}{1 + \frac{h}{R}}$$
At maximum height, $v = 0$, so $\Delta U = K_i$:
$$\frac{m g_0 h}{1 + \frac{h}{R}} = \frac{1}{4} m g_0 R \implies \frac{h / R}{1 + h / R} = \frac{1}{4}$$
$$4\left(\frac{h}{R}\right) = 1 + \frac{h}{R} \implies 3\left(\frac{h}{R}\right) = 1 \implies \mathbf{h = \frac{R}{3}}$$
Correct Answer: (B)
Solution 3:
The escape velocity is given by $v_e = \sqrt{\frac{2 G M}{R}}$.
Since $v_e$ depends only on the mass $M$ and radius $R$ of the planet and is completely independent of the mass $m$ of the projectile, the escape velocity remains identical at $\mathbf{11.2\text{ km/s}}$.
Correct Answer: (B)
Solution 4:
At altitude $h = 3R$, the distance from the center of Earth is $r = R + h = R + 3R = 4R$.
The escape velocity from this altitude is:
$$v_e’ = \sqrt{\frac{2 G M}{r}} = \sqrt{\frac{2 G M}{4 R}} = \frac{1}{2} \sqrt{\frac{2 G M}{R}} = \mathbf{\frac{v_e}{2}}$$
Correct Answer: (A)
Solution 5:
Escape velocity is derived from scalar energy conservation ($E = K + U = 0$). As long as the velocity vector is directed above the horizon (so that the trajectory does not collide with the Earth’s surface), the required launch speed is completely independent of the projection angle. Hence, $v = \mathbf{11.2\text{ km/s}}$.
Correct Answer: (A)
Solution 6:
– (A) True: $V_{\text{in}}(R) = V_{\text{out}}(R) = -G M / R$ (continuous).
– (B) True: $E_{\text{in}}(R) = E_{\text{out}}(R) = -G M / R^2$ (continuous).
– (C) True: $V(r) = -\frac{G M}{2 R^3}(3 R^2 – r^2)$ has its minimum at $r = 0$, where $V_c = -1.5 G M / R$.
– (D) False: Potential is quadratic in $r$ inside, not constant. (It is constant only inside a hollow shell).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 7:
– (A) True: Total energy $E = \frac{1}{2}mv^2 – G M m/R > 0$; orbits with $E > 0$ are hyperbolic.
– (B) True: $\frac{1}{2}mv_\infty^2 = \frac{1}{2}mv^2 – \frac{1}{2}mv_e^2 \implies v_\infty = \sqrt{v^2 – v_e^2}$.
– (C) True: Positive total energy characterizes unbounded hyperbolic states.
– (D) False: Gravity continues to decelerate the rocket throughout its outward flight ($a < 0$).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 8:
– (A) Setting $v_e = c$: $c = \sqrt{\frac{2 G M}{R_s}} \implies R_s = \frac{2 G M}{c^2}$. (True).
– (B) For Earth: $R_s = \frac{2 (6.674 \times 10^{-11})(6.0 \times 10^{24})}{(3.0 \times 10^8)^2} = \frac{8.0 \times 10^{14}}{9.0 \times 10^{16}} \approx 8.9 \times 10^{-3}\text{ m} \approx 9\text{ mm}$. (True).
– (C) Volume $V \propto R_s^3 \propto M^3$. Density $\rho = M / V \propto M / M^3 = M^{-2}$. (True).
– (D) The event horizon $R_s$ defines the boundary of no return. (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
By energy conservation:
$$v_\infty = \sqrt{v^2 – v_e^2} = \sqrt{(3 v_e)^2 – v_e^2} = \sqrt{9 v_e^2 – v_e^2} = \sqrt{8} v_e$$
Comparing with $v_\infty = \sqrt{n} v_e$:
$$\mathbf{n = 8}$$
Correct Answer: 8
Solution 10:
Escape velocity from the center of Earth:
$$v_{e, \text{center}} = \sqrt{3 g_0 R}$$
Escape velocity from the surface of Earth:
$$v_{e, \text{surface}} = \sqrt{2 g_0 R}$$
Taking the ratio:
$$\frac{v_{e, \text{center}}}{v_{e, \text{surface}}} = \frac{\sqrt{3 g_0 R}}{\sqrt{2 g_0 R}} = \sqrt{\frac{3}{2}}$$
Comparing with $\sqrt{\frac{k}{2}}$:
$$\mathbf{k = 3}$$
Correct Answer: 3