**Galileo** (1564–1642) established that all bodies, regardless of mass, accelerate towards Earth with constant acceleration due to gravity. His experiments with inclined planes provided early understanding of gravitational acceleration.
**Historical Models of Planetary Motion:**
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**Kepler's laws describe the motion of planets around the Sun with mathematical elegance and form the foundation for Newton's law of gravitation.**
**Statement:** All planets move in elliptical orbits with the Sun situated at one of the two foci of the ellipse.
**Understanding Ellipses:**
**Statement:** The line (radius vector) joining a planet to the Sun sweeps out equal areas in equal intervals of time.
**Mathematical Expression:**
$$\frac{\Delta A}{\Delta t} = \text{constant}$$
**Derivation from Angular Momentum Conservation:**
For a planet of mass m at position **r** with velocity **v**, the area swept in time Δt is:
$$\Delta A = \frac{1}{2}|\mathbf{r} \times \mathbf{v}\Delta t|$$
$$\frac{\Delta A}{\Delta t} = \frac{1}{2}|\mathbf{r} \times \mathbf{v}| = \frac{L}{2m}$$
where **L** = **r** × **p** = **r** × m**v** is angular momentum.
Since gravitation is a **central force** (directed along **r**), angular momentum L is conserved:
$$\frac{\Delta A}{\Delta t} = \text{constant}$$
**Physical Consequence:** Planets move faster near perihelion (closer to Sun) and slower near aphelion (farther from Sun).
**Example:** For a planet, if rₚ is perihelion distance and vₚ is speed at perihelion, and rₐ is aphelion distance with vₐ speed at aphelion, then:
$$r_p v_p = r_a v_a$$
Since rₐ > rₚ, we have vₚ > vₐ.
**Statement:** The square of the period of revolution T of a planet is directly proportional to the cube of the semi-major axis a of its elliptical orbit.
**Mathematical Expression:**
$$T^2 \propto a^3$$
or
$$\frac{T^2}{a^3} = \text{constant}$$
**Verification:** Table 7.1 in NCERT shows that for all planets, the quotient T²/a³ ≈ 2.96–3.01 × 10⁻³⁴ y²/m³, confirming the law.
**SI Units:**
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**Historical Context:** Newton observed an apple falling from a tree and reasoned that the same gravitational force holding the apple to Earth also keeps the Moon in orbit around Earth.
**Calculation of Moon's Acceleration:**
The Moon orbits Earth in circular motion with:
Centripetal acceleration:
$$a_m = \frac{V^2}{R_m} = \frac{4\pi^2 R_m}{T^2} \approx 0.0027 \text{ m/s}^2$$
This is much less than g = 9.8 m/s² at Earth's surface, suggesting gravitational force decreases with distance.
**Inverse-Square Law Hypothesis:**
If gravitational force ∝ 1/r², then:
$$\frac{g}{a_m} = \frac{R_m^2}{R_E^2} = \frac{(3.84 \times 10^8)^2}{(6.37 \times 10^6)^2} \approx 3600$$
This agrees with observations, confirming the inverse-square law.
**Statement:** Every body in the universe attracts every other body with a force that is:
1. **Directly proportional** to the product of their masses
2. **Inversely proportional** to the square of the distance between them
**Mathematical Expression (Scalar):**
$$F = G\frac{m_1 m_2}{r^2}$$
where:
**Vector Form:**
$$\mathbf{F}_{12} = -G\frac{m_1 m_2}{r^2}\hat{\mathbf{r}}$$
or
$$\mathbf{F}_{12} = -G\frac{m_1 m_2}{r^3}\mathbf{r}$$
where:
**Key Points:**
The law strictly applies to point masses. For extended objects, we must sum forces vectorially from all mass elements.
**Principle of Superposition:**
For multiple masses m₂, m₃, m₄, ... acting on mass m₁:
$$\mathbf{F}_{total} = \mathbf{F}_{21} + \mathbf{F}_{31} + \mathbf{F}_{41} + ...$$
$$\mathbf{F}_{total} = -Gm_1\left(\frac{m_2}{r_{21}^2}\hat{\mathbf{r}}_{21} + \frac{m_3}{r_{31}^2}\hat{\mathbf{r}}_{31} + ...\right)$$
**Example 7.2:** Three equal masses m at vertices of equilateral triangle with centroid G.
**Part (a) Force on mass 2m at G:**
By symmetry, forces from the three masses are separated by 120° angles and are equal in magnitude. Their vector sum is **zero**: **F_R** = 0.
**Part (b) If mass at A is doubled to 2m:**
Force from mass at A has magnitude: F_A = G(2m)(2m)/1² = 4Gm²
Forces from masses at B and C each: F_B = F_C = Gm(2m)/1² = 2Gm²
The resultant is now non-zero, directed from G towards A:
$$F_R = 4Gm^2 - 2Gm^2\cos(60°) = 3Gm^2$$
For calculating gravitational forces of extended objects, two critical theorems apply:
**Theorem 1 - Shell Outside Point:**
A hollow spherical shell of uniform density exerts a gravitational force on an external point mass as if the entire mass of the shell is concentrated at its centre.
**Physical Explanation:** Gravitational forces from different parts of the shell have perpendicular components that cancel; only the radial component remains.
**Theorem 2 - Shell Inside Point:**
A hollow spherical shell of uniform density exerts **zero net gravitational force** on a point mass situated inside it.
**Physical Explanation:** Forces from diametrically opposite regions cancel completely.
**Consequence for Uniform Sphere:**
---
**Objective:** First experimental measurement of G, the universal gravitational constant.
**Apparatus:**
**Principle:** The gravitational attraction between each large and small sphere creates a torque that twists the wire. At equilibrium, gravitational torque equals restoring torque of the wire.
**Working:**
For each pair of spheres (large mass M at distance d from small mass m):
$$F = G\frac{Mm}{d^2}$$
If L is the length of bar AB, the gravitational torque is:
$$\tau_{grav} = F \cdot L = G\frac{MmL}{d^2}$$
The wire's restoring torque is:
$$\tau_{restore} = \kappa\theta$$
where κ is the restoring couple per unit angle of twist and θ is the angle of rotation.
At equilibrium:
$$G\frac{MmL}{d^2} = \kappa\theta$$
$$G = \frac{\kappa\theta d^2}{MmL}$$
By independently measuring κ (applying known torque and measuring rotation) and observing θ, we can calculate G.
**Currently Accepted Value:**
$$\boxed{G = 6.67 \times 10^{-11} \text{ N·m}^2\text{/kg}^2}$$
**Significance:** This constant is universal—same throughout the universe, independent of the nature of the two bodies or the medium between them.
---
**Model:** Consider Earth as a sphere of uniform density ρ, mass M_E, and radius R_E.
For a point mass m on Earth's surface:
By the shell theorem, Earth acts as if its entire mass is concentrated at the centre.
Using Newton's second law with gravitational force:
$$F = G\frac{M_E m}{R_E^2} = mg$$
$$\boxed{g = \frac{GM_E}{R_E^2}}$$
**SI Units:** g in m/s²
**Numerical Value:** g ≈ 9.8 m/s² (exact value depends on location; varies from 9.78 m/s² at equator to 9.83 m/s² at poles)
**Determination of Earth's Mass:**
From the equation above:
$$M_E = \frac{gR_E^2}{G} = \frac{9.8 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} \approx 5.97 \times 10^{24} \text{ kg}$$
This is why **Cavendish's determination of G is called "weighing the Earth"**—it allowed calculation of Earth's mass.
**Assumptions:**
---
**Scenario:** Point mass m at height h above Earth's surface.
Distance from Earth's centre: r = R_E + h
**Force and Acceleration:**
$$F(h) = G\frac{M_E m}{(R_E + h)^2}$$
$$g(h) = \frac{F(h)}{m} = \frac{GM_E}{(R_E + h)^2}$$
$$\boxed{g(h) = \frac{GM_E}{(R_E + h)^2}}$$
**Relation to Surface Gravity:**
Since g = GM_E/R_E²:
$$g(h) = g\left(\frac{R_E}{R_E + h}\right)^2 = g\left(\frac{1}{1 + h/R_E}\right)^2$$
**Binomial Expansion:**
Using $(1 + x)^{-2} \approx 1 - 2x$ for |x| << 1:
$$g(h) = g\left(1 + \frac{h}{R_E}\right)^{-2} \approx g\left(1 - \frac{2h}{R_E}\right)$$
$$\boxed{g(h) \approx g\left(1 - \frac{2h}{R_E}\right)}$$
**Interpretation:** For every kilometre of height, g decreases by approximately 2/R_E = 2/(6.37 × 10⁶) ≈ 3.1 × 10⁻⁷ (fractional decrease).
**Example Calculation:**
At h = 100 m above surface:
$$g(100) \approx 9.8\left(1 - \frac{2 \times 100}{6.37 \times 10^6}\right) = 9.8(1 - 3.14 \times 10^{-5}) \approx 9.8 \text{ m/s}^2$$
The change is negligible for typical altitudes.
---
**Scenario:** Point mass m at depth d below Earth's surface.
Distance from Earth's centre: r = R_E − d
**Shell Model:**
Earth can be divided into:
1. **Inner sphere** of radius (R_E − d) containing mass M_s
2. **Outer spherical shell** of thickness d and mass (M_E − M_s)
By the shell theorem:
**Mass of Inner Sphere:**
Assuming uniform density:
$$\frac{M_s}{M_E} = \left(\frac{R_E - d}{R_E}\right)^3$$
$$M_s = M_E\left(\frac{R_E - d}{R_E}\right)^3$$
**Gravitational Force and Acceleration:**
$$F(d) = G\frac{M_s m}{(R_E - d)^2} = G\frac{M_E m}{R_E^3}(R_E - d)$$
$$g(d) = \frac{GM_E}{R_E^3}(R_E - d)$$
$$\boxed{g(d) = g\left(1 - \frac{d}{R_E}\right)}$$
or equivalently:
$$\boxed{g(d) = g\frac{R_E - d}{R_E}}$$
**Key Observations:**
1. **g decreases linearly with depth** (unlike the inverse-square dependence above the surface)
2. At Earth's surface (d = 0): g(0) = g
3. At Earth's centre (d = R_E): g(R_E) = 0 (no net force at centre)
4. The linear relationship assumes uniform density
**Example Calculation:**
At d = 3,200 km (halfway to Earth's centre):
$$g = 9.8\left(1 - \frac{3.2 \times 10^6}{6.37 \times 10^6}\right) = 9.8(1 - 0.50) \approx 4.9 \text{ m/s}^2$$
**Physical Interpretation:** Although you are closer to Earth's mass at greater depths, you are surrounded by a shell of Earth that exerts zero net force, and the effective mass pulling you downward decreases linearly.
---
**Gravitational Potential Energy (U):** The energy possessed by a body due to its position in a gravitational field.
**Derivation:** Consider a point mass m at distance r from a mass M (e.g., Earth at origin).
Work done by external agent in moving m from r₁ to r₂ against gravity:
$$W_{external} = -\int_{r_1}^{r_2} F_{grav} dr = -\int_{r_1}^{r_2} \left(-G\frac{Mm}{r^2}\right) dr$$
$$W_{external} = GM m\int_{r_1}^{r_2} \frac{dr}{r^2} = GMm\left[-\frac{1}{r}\right]_{r_1}^{r_2}$$
$$W_{external} = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$
This work equals the change in potential energy:
$$U(r_2) - U(r_1) = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$
By convention, setting U(∞) = 0 (potential energy is zero at infinity):
$$\boxed{U(r) = -\frac{GMm}{r}}$$
**Key Features:**
For a body of mass m at height h above Earth's surface:
$$U(h) = -\frac{GM_E m}{R_E + h}$$
For small heights (h << R_E):
$$U(h) = -\frac{GM_E m}{R_E}\left(1 + \frac{h}{R_E}\right)^{-1} \approx -\frac{GM_E m}{R_E}\left(1 - \frac{h}{R_E}\right)$$
$$U(h) \approx -\frac{GM_E m}{R_E} + \frac{GM_E m h}{R_E^2}$$
Since g = GM_E/R_E²:
$$U(h) \approx U_0 + mgh$$
where U₀ = −GM_Em/R_E is the potential energy at Earth's surface.
**For practical problems near Earth's surface**, we use the approximation:
$$\boxed{U(h) = mgh \text{ (taking surface as reference: } U(0) = 0\text{)}}$$
This is valid only when h is small compared to R_E and acceleration g is approximately constant.
---
**Escape Speed (or Escape Velocity):** The minimum speed required for a body to escape from the gravitational field of a celestial body, reaching infinity with zero velocity.
**Energy Conservation Approach:**
For a body of mass m at Earth's surface with initial speed v_e:
Total mechanical energy = Kinetic energy + Potential energy
$$E = \frac{1}{2}mv_e^2 + \left(-\frac{GM_E m}{R_E}\right)$$
For the body to escape to infinity with zero velocity:
$$E_{final} = 0 + 0 = 0$$
By conservation of energy:
$$\frac{1}{2}mv_e^2 - \frac{GM_E m}{R_E} = 0$$
$$\frac{1}{2}v_e^2 = \frac{GM_E}{R_E}$$
$$\boxed{v_e = \sqrt{\frac{2GM_E}{R_E}}}$$
**Relation to Surface Gravity:**
Since g = GM_E/R_E²:
$$v_e = \sqrt{2gR_E}$$
$$\boxed{v_e = \sqrt{2gR_E}}$$
**Numerical Calculation for Earth:**
$$v_e = \sqrt{2 \times 9.8 \times 6.37 \times 10^6} = \sqrt{1.25 \times 10^8} \approx 11.2 \text{ km/s}$$
**Key Points:**
**Escape Speeds for Other Bodies:**
| Body | v_e (km/s) |
|------|-----------|
| Moon | 2.4 |
| Mercury | 4.3 |
| Venus | 10.4 |
| Mars | 5.0 |
| Jupiter | 59.5 |
| Sun | 618 |
**Physical Interpretation:** The escape speed represents the threshold beyond which gravitational attraction can no longer pull back the escaping body.
---
**Satellite:** A natural or artificial body orbiting a larger celestial body under gravitational attraction.
**Condition for Circular Orbit:**
For a satellite of mass m in circular orbit at height h above Earth's surface (orbital radius r = R_E + h):
The gravitational force provides the centripetal force required for circular motion:
$$\frac{GM_E m}{r^2} = \frac{mv_{orbit}^2}{r}$$
$$v_{orbit} = \sqrt{\frac{GM_E}{r}}$$
$$\boxed{v_{orbit} = \sqrt{\frac{GM_E}{R_E + h}}}$$
For a satellite on Earth's surface (h = 0):
$$v_{orbit} = \sqrt{\frac{GM_E}{R_E}} = \sqrt{gR_E} \approx 7.9 \text{ km/s}$$
**Period of Revolution:**
Orbital speed is also: $v_{orbit} = \frac{2\pi r}{T}$
where T is the period of revolution.
$$\frac{2\pi r}{T} = \sqrt{\frac{GM_E}{r}}$$
$$T = 2\pi r\sqrt{\frac{r}{GM_E}} = 2\pi\sqrt{\frac{r^3}{GM_E}}$$
$$\boxed{T = 2\pi\sqrt{\frac{(R_E + h)^3}{GM_E}}}$$
Alternatively:
$$\boxed{T = 2\pi\sqrt{\frac{r^3}{GM_E}}}$$
**Kepler's Third Law Verification:**
Squaring the period equation:
$$T^2 = \frac{4\pi^2 r^3}{GM_E}$$
$$\frac{T^2}{r^3} = \frac{4\pi^2}{GM_E} = \text{constant}$$
This is Kepler's third law, confirming that T² ∝ r³.
**Definition:** A satellite with orbital period equal to Earth's rotational period (T = 24 hours), appearing stationary above a fixed point on the equator.
**Orbital Radius for Geostationary Satellite:**
Period: T = 24 hours = 86,400 s
From $T^2 = \frac{4\pi^2 r^3}{GM_E}$:
$$r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2}$$
$$r^3 \approx 7.53 \times 10^{22} \text{ m}^3$$
$$r \approx 4.22 \times 10^7 \text{ m} = 42,200 \text{ km}$$
Height above Earth's surface:
$$h = r - R_E = 42,200 - 6,371 \approx 35,800 \text{ km}$$
**Applications:**
**Orbital Speed:**
$$v_{orbit} = \sqrt{\frac{GM_E}{r}} = \frac{2\pi r}{T} \approx 3.1 \text{ km/s}$$
**Angular Momentum Conservation:**
For satellite in circular orbit:
$$L = mvr = m\sqrt{\frac{GM_E}{r}} \cdot r = m\sqrt{GM_E r}$$
**Centripetal Acceleration:**
$$a_c = \frac{v_{orbit}^2}{r} = \frac{GM_E}{r^2} = g(r)$$
The gravitational acceleration at orbital radius r provides the exact centripetal acceleration needed.
---
**Kinetic Energy (KE):**
For satellite in circular orbit with speed $v = \sqrt{\frac{GM_E}{r}}$:
$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM_E}{r} = \frac{GM_E m}{2r}$$
**Potential Energy (PE):**
$$PE = -\frac{GM_E m}{r}$$
**Total Energy:**
$$E = KE + PE = \frac{GM_E m}{2r} - \frac{GM_E m}{r}$$
$$\boxed{E = -\frac{GM_E m}{2r}}$$
**Key Observation:**
The total energy is **negative**, indicating that the satellite is in a **bound orbit** and cannot escape to infinity without external work.
**Relation Between Energies:**
$$KE = -\frac{E}{1} = \frac{GM_E m}{2r}$$
$$PE = 2E = -\frac{GM_E m}{r}$$
$$KE = -PE/2 \quad \text{(Virial Theorem)}$$
**Binding Energy:** The energy required to move a satellite from its orbit to infinity (to escape).
$$E_{binding} = -E = \frac{GM_E m}{2r}$$
**Energy to Move from Orbit r₁ to Orbit r₂:**
$$\Delta E = E_2 - E_1 = -\frac{GM_E m}{2r_2} + \frac{GM_E m}{2r_1}$$
$$\Delta E = \frac{GM_E m}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$
If r₂ > r₁ (higher orbit), then ΔE > 0, and external energy must be supplied.
From $E = -\frac{GM_E m}{2r}$ and $E = \frac{1}{2}mv^2 - \frac{GM_E m}{r}$:
$$\frac{1}{2}mv^2 - \frac{GM_E m}{r} = -\frac{GM_E m}{2r}$$
$$\frac{1}{2}mv^2 = \frac{GM_E m}{r} - \frac{GM_E m}{2r} = \frac{GM_E m}{2r}$$
$$v = \sqrt{\frac{GM_E}{r}}$$
This confirms the orbital speed formula.
A satellite at orbital radius r has speed $v_{orbit} = \sqrt{\frac{GM_E}{r}}$.
To escape from this orbit to infinity:
$$v_{escape} = \sqrt{\frac{2GM_E}{r}} = \sqrt{2} \cdot v_{orbit}$$
The escape speed from any orbit is **√2 times the orbital speed** at that radius.
**Example:** For a geostationary satellite:
---
$$F = G\frac{m_1 m_2}{r^2}$$
**Dimensional Check:**
$$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{MLT^{-2} \cdot L^2}{M \cdot M} = M^{-1}L^3T^{-2}$$
**SI Units:** N·m²·kg⁻² or m³·kg⁻¹·s⁻²
$$U = -\frac{GMm}{r}$$
**Dimensional Check:**
$$[U] = [G][M][m]/[r] = (M^{-1}L^3T^{-2})(M)(M)/L = ML^2T^{-2}$$
This is energy (Joules).
$$v = \sqrt{\frac{GM}{r}}$$
**Dimensional Check:**
$$[v] = \sqrt{\frac{[G][M]}{[r]}} = \sqrt{\frac{(M^{-1}L^3T^{-2})(M)}{L}} = \sqrt{L^2T^{-2}} = LT^{-1}$$
This is velocity (m/s).
$$v_e = \sqrt{2gR}$$
**Dimensional Check:**
$$[v_e] = \sqrt{[g][R]} = \sqrt{(LT^{-2})(L)} = \sqrt{L^2T^{-2}} = LT^{-1}$$
Correct dimensions for velocity.
---
**Kepler's Laws:**
**Universal Law of Gravitation:**
**Gravitational Field Strength:**
**Acceleration at Height h:**
**Acceleration at Depth d:**
**Gravitational Potential Energy:**
**Escape Speed:**
**Orbital Speed:**
**Orbital Period:**
Q1. According to Kepler's first law, planets move in orbits that are:
Answer: B — Kepler's first law explicitly states that planetary orbits are elliptical with the Sun at one focus, not circles as Copernicus proposed.
Q2. If a planet is at perihelion (closest to Sun) with orbital speed v_P and distance r_P, and at aphelion (farthest) with speed v_A and distance r_A, which relation is correct?
Answer: A — Angular momentum L = mvr is conserved for a central force; hence m·v_P·r_P = m·v_A·r_A, giving v_P·r_P = v_A·r_A.
Q3. The gravitational constant G in SI units is numerically equal to:
Answer: B — G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant; g ≈ 9.8 m/s² is acceleration due to gravity at Earth's surface.
Q4. A satellite orbits Earth at height h = R (where R is Earth's radius). What is its orbital speed compared to the orbital speed at Earth's surface?
Answer: B — Orbital speed v_o = √(GM/r); at height h = R, orbital radius is r = 2R, so v = √(GM/2R) = √(GM/R)/√2 = v_surface/√2.
Q5. Kepler's second law (law of areas) is a direct consequence of which principle?
Answer: B — Since gravitational force is a central force with zero torque about the Sun, angular momentum is conserved, leading to constant areal velocity (law of areas).
Q6. If the escape speed from Earth's surface is v_e = 11.2 km/s and a planet has twice Earth's mass and twice Earth's radius, the escape speed from that planet's surface is approximately:
Answer: B — v_e = √(2GM/R); if M → 2M and R → 2R, then v_e(new) = √(2G·2M/2R) = √(2GM/R) = v_e(original) ≈ 11.2 km/s.
Q7. Assertion: Acceleration due to gravity is greater at the equator than at the poles. Reason: Earth's equator is closer to the centre than the poles.
Answer: D — g is actually greater at poles and smaller at equator due to Earth's rotation (centrifugal effect) and oblate shape; the equator is farther from centre, reducing g further.
Q8. A geostationary satellite must have an orbital period equal to Earth's rotation period (24 hours). Its orbital radius from Earth's centre is approximately:
Answer: C — Using T² = (4π²/GM)r³ with T = 24 hours, r ≈ 42,000 km from Earth's centre (or ~36,000 km altitude above surface).
Q9. The total mechanical energy of a satellite in circular orbit at radius r is E = −GMm/(2r). Which statement is NOT correct?
Answer: D — As a satellite moves to a lower orbit (smaller r), orbital speed v_o = √(GM/r) increases, not decreases; all other statements are correct.
Q10. Derive that for a planet in circular orbit, the orbital speed v_o = √(GM/r), starting from Newton's law of gravitation and centripetal force requirement. [HOTS]
Answer: A — For circular orbit, gravitational force provides centripetal force: GMm/r² = mv_o²/r; solving for v_o gives v_o = √(GM/r).
What is the mathematical statement of Kepler's first law?
All planets move in elliptical orbits with the Sun located at one of the two foci of the ellipse.
State Kepler's second law (law of areas) in one sentence.
The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
Write Kepler's third law as a mathematical equation.
T² is proportional to a³, or T² = ka³ where k is a constant and a is the semi-major axis.
What is the universal law of gravitation formula?
F = GMm/r², where G is the gravitational constant, M and m are masses, and r is the distance between them.
What is the numerical value and SI unit of the gravitational constant G?
G = 6.67 × 10⁻¹¹ N·m²/kg² (or m³·kg⁻¹·s⁻²).
How does acceleration due to gravity g change with altitude h above Earth's surface?
g decreases with altitude: g_h = g₀(R/(R+h))² or approximately g_h ≈ g₀(1 − 2h/R) for small h.
Define escape speed and write its formula.
Escape speed is the minimum speed needed for an object to leave Earth's gravitational field permanently; v_e = √(2GM/R).
What is the relationship between orbital speed, orbital radius, and gravitational force?
Orbital speed v_o = √(GM/r); the gravitational force provides the centripetal force needed for circular motion.
Why is angular momentum conserved for a planet orbiting the Sun?
Gravitational force is a central force directed along the radius vector, so the torque about the Sun is zero, keeping angular momentum constant.
What is the total mechanical energy of a satellite in a circular orbit?
Total energy E = KE + PE = ½mv_o² − GMm/r = −GMm/(2r), which is negative, indicating a bound orbit.
State Kepler's law of periods and verify it using the data provided in Table 7.1 for any two planets. [2 marks]
State T² ∝ a³ or T² = Ka³; calculate T²/a³ for two planets from Table 7.1 and show both give approximately the same value (≈ 3.0 × 10⁻³⁴ y²/m³).
Derive the expression for acceleration due to gravity at a height h above Earth's surface, starting from the universal law of gravitation. Show that g_h = g₀[R/(R+h)]² and for small h, g_h ≈ g₀(1 − 2h/R). [5 marks]
At surface: g₀ = GM/R². At height h: g_h = GM/(R+h)². Divide to get ratio; use binomial approximation (1+x)ⁿ ≈ 1 + nx for |x| << 1 with x = h/R and n = −2.
A satellite of mass m orbits Earth at radius r. Derive expressions for (a) orbital speed v_o, (b) orbital period T, (c) kinetic energy KE, (d) potential energy PE, and (e) total energy E. Show that KE = −E and explain why the total energy is negative. [HOTS] [6 marks]
Set gravitational force = centripetal force to find v_o = √(GM/r). Use T = 2πr/v_o. Calculate KE = ½mv_o² = GMm/(2r) and PE = −GMm/r. Total E = KE + PE = −GMm/(2r). Negative E means bound orbit; E > 0 would mean escape.
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