Top 50 Most Repeated GRAVITATION PYQs | JEE MAINS
A curated collection of the most important questions from GRAVITATION, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from GRAVITATION, fully solved with step-by-step concepts to prepare for JEE MAINS.
Kepler's third law relates orbital period squared to semi-major axis cubed....
Read Full Step-by-Step Solution →Using $T = 24\,\text{hr}$, $\frac{GMm}{r^2} = m\omega^2 r$, solve $r \approx 6.6R \to 7$ (nearest integer)....
Read Full Step-by-Step Solution →By definition, the gravitational field is the negative gradient of the potential: $\mathbf{g}= -\nabla V$. In one dimension this reduces to $g = -\fra...
Read Full Step-by-Step Solution →Orbital speed $v = 2\pi r / T$. Convert period: $90\,\text{min}=5400\,\text{s}$. Then $v = 2\pi(7.0\times10^{6})/5400 \approx 8.15\times10^{3}\,\text{...
Read Full Step-by-Step Solution →Escape velocity is derived from energy conservation: $\frac{1}{2}mv^2 = \frac{GMm}{R}$....
Read Full Step-by-Step Solution →Escape velocity is given by \(v = \sqrt{\frac{2GM}{R}}\). Substituting the constants yields \(v \approx 1.12\times10^{4}\) m s⁻¹ = 11.2 km s⁻¹....
Read Full Step-by-Step Solution →Use $g' = g\left(\frac{R}{R+h}\right)^2$. Substituting $R=6.4\times10^6\ \text{m}$ and $h=10^4\ \text{m}$ gives $g'\approx9.8\left(\frac{6.4\times10^6...
Read Full Step-by-Step Solution →F = G m₁ m₂ / r² = (6.674×10⁻¹¹)(2×3) / (0.5)² = 1.60×10⁻⁹ N....
Read Full Step-by-Step Solution →It must match Earth's rotational period to appear stationary relative to the ground....
Read Full Step-by-Step Solution →Orbital speed v = 2πr / T. Substituting r = 7.0×10⁶ m and T = 5400 s gives v ≈ 8.1×10³ m/s = 8.1 km/s....
Read Full Step-by-Step Solution →Treating the orbit as circular allows the use of centripetal force = gravitational force, leading to T² ∝ r³, which is Kepler's third law....
Read Full Step-by-Step Solution →Using T = 2π√(r³/GM), solve for r: r = (GMT²/4π²)^{1/3}. With T = 86 400 s, r ≈ 4.22×10⁷ m. Altitude = r – R_E ≈ 4.22×10⁷ − 6.37×10⁶ ≈ 3.58×10⁷ m = 35...
Read Full Step-by-Step Solution →Foundational check for Gravitation. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Kepler's third law relates orbital period squared to semi-major axis cubed....
Read Full Step-by-Step Solution →Using $T = 24\,\text{hr}$, $\frac{GMm}{r^2} = m\omega^2 r$, solve $r \approx 6.6R \to 7$ (nearest integer)....
Read Full Step-by-Step Solution →By definition, the gravitational field is the negative gradient of the potential: $\mathbf{g}= -\nabla V$. In one dimension this reduces to $g = -\fra...
Read Full Step-by-Step Solution →Orbital speed $v = 2\pi r / T$. Convert period: $90\,\text{min}=5400\,\text{s}$. Then $v = 2\pi(7.0\times10^{6})/5400 \approx 8.15\times10^{3}\,\text{...
Read Full Step-by-Step Solution →Escape velocity is derived from energy conservation: $\frac{1}{2}mv^2 = \frac{GMm}{R}$....
Read Full Step-by-Step Solution →Escape velocity is given by \(v = \sqrt{\frac{2GM}{R}}\). Substituting the constants yields \(v \approx 1.12\times10^{4}\) m s⁻¹ = 11.2 km s⁻¹....
Read Full Step-by-Step Solution →Use $g' = g\left(\frac{R}{R+h}\right)^2$. Substituting $R=6.4\times10^6\ \text{m}$ and $h=10^4\ \text{m}$ gives $g'\approx9.8\left(\frac{6.4\times10^6...
Read Full Step-by-Step Solution →F = G m₁ m₂ / r² = (6.674×10⁻¹¹)(2×3) / (0.5)² = 1.60×10⁻⁹ N....
Read Full Step-by-Step Solution →It must match Earth's rotational period to appear stationary relative to the ground....
Read Full Step-by-Step Solution →Orbital speed v = 2πr / T. Substituting r = 7.0×10⁶ m and T = 5400 s gives v ≈ 8.1×10³ m/s = 8.1 km/s....
Read Full Step-by-Step Solution →Treating the orbit as circular allows the use of centripetal force = gravitational force, leading to T² ∝ r³, which is Kepler's third law....
Read Full Step-by-Step Solution →Using T = 2π√(r³/GM), solve for r: r = (GMT²/4π²)^{1/3}. With T = 86 400 s, r ≈ 4.22×10⁷ m. Altitude = r – R_E ≈ 4.22×10⁷ − 6.37×10⁶ ≈ 3.58×10⁷ m = 35...
Read Full Step-by-Step Solution →Foundational check for Gravitation. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Kepler's third law relates orbital period squared to semi-major axis cubed....
Read Full Step-by-Step Solution →Using $T = 24\,\text{hr}$, $\frac{GMm}{r^2} = m\omega^2 r$, solve $r \approx 6.6R \to 7$ (nearest integer)....
Read Full Step-by-Step Solution →By definition, the gravitational field is the negative gradient of the potential: $\mathbf{g}= -\nabla V$. In one dimension this reduces to $g = -\fra...
Read Full Step-by-Step Solution →Orbital speed $v = 2\pi r / T$. Convert period: $90\,\text{min}=5400\,\text{s}$. Then $v = 2\pi(7.0\times10^{6})/5400 \approx 8.15\times10^{3}\,\text{...
Read Full Step-by-Step Solution →Escape velocity is derived from energy conservation: $\frac{1}{2}mv^2 = \frac{GMm}{R}$....
Read Full Step-by-Step Solution →Escape velocity is given by \(v = \sqrt{\frac{2GM}{R}}\). Substituting the constants yields \(v \approx 1.12\times10^{4}\) m s⁻¹ = 11.2 km s⁻¹....
Read Full Step-by-Step Solution →Use $g' = g\left(\frac{R}{R+h}\right)^2$. Substituting $R=6.4\times10^6\ \text{m}$ and $h=10^4\ \text{m}$ gives $g'\approx9.8\left(\frac{6.4\times10^6...
Read Full Step-by-Step Solution →F = G m₁ m₂ / r² = (6.674×10⁻¹¹)(2×3) / (0.5)² = 1.60×10⁻⁹ N....
Read Full Step-by-Step Solution →It must match Earth's rotational period to appear stationary relative to the ground....
Read Full Step-by-Step Solution →Orbital speed v = 2πr / T. Substituting r = 7.0×10⁶ m and T = 5400 s gives v ≈ 8.1×10³ m/s = 8.1 km/s....
Read Full Step-by-Step Solution →Treating the orbit as circular allows the use of centripetal force = gravitational force, leading to T² ∝ r³, which is Kepler's third law....
Read Full Step-by-Step Solution →Using T = 2π√(r³/GM), solve for r: r = (GMT²/4π²)^{1/3}. With T = 86 400 s, r ≈ 4.22×10⁷ m. Altitude = r – R_E ≈ 4.22×10⁷ − 6.37×10⁶ ≈ 3.58×10⁷ m = 35...
Read Full Step-by-Step Solution →Foundational check for Gravitation. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →Kepler's third law relates orbital period squared to semi-major axis cubed....
Read Full Step-by-Step Solution →Using $T = 24\,\text{hr}$, $\frac{GMm}{r^2} = m\omega^2 r$, solve $r \approx 6.6R \to 7$ (nearest integer)....
Read Full Step-by-Step Solution →By definition, the gravitational field is the negative gradient of the potential: $\mathbf{g}= -\nabla V$. In one dimension this reduces to $g = -\fra...
Read Full Step-by-Step Solution →Orbital speed $v = 2\pi r / T$. Convert period: $90\,\text{min}=5400\,\text{s}$. Then $v = 2\pi(7.0\times10^{6})/5400 \approx 8.15\times10^{3}\,\text{...
Read Full Step-by-Step Solution →Escape velocity is derived from energy conservation: $\frac{1}{2}mv^2 = \frac{GMm}{R}$....
Read Full Step-by-Step Solution →Escape velocity is given by \(v = \sqrt{\frac{2GM}{R}}\). Substituting the constants yields \(v \approx 1.12\times10^{4}\) m s⁻¹ = 11.2 km s⁻¹....
Read Full Step-by-Step Solution →Use $g' = g\left(\frac{R}{R+h}\right)^2$. Substituting $R=6.4\times10^6\ \text{m}$ and $h=10^4\ \text{m}$ gives $g'\approx9.8\left(\frac{6.4\times10^6...
Read Full Step-by-Step Solution →F = G m₁ m₂ / r² = (6.674×10⁻¹¹)(2×3) / (0.5)² = 1.60×10⁻⁹ N....
Read Full Step-by-Step Solution →It must match Earth's rotational period to appear stationary relative to the ground....
Read Full Step-by-Step Solution →Orbital speed v = 2πr / T. Substituting r = 7.0×10⁶ m and T = 5400 s gives v ≈ 8.1×10³ m/s = 8.1 km/s....
Read Full Step-by-Step Solution →Treating the orbit as circular allows the use of centripetal force = gravitational force, leading to T² ∝ r³, which is Kepler's third law....
Read Full Step-by-Step Solution →