Therefore, the required distance is √(5)R - R2$\frac{\sqrt{5}R - R}{2}$.
Pattern Recognition
Do not use the linear approximation formula gh ≈ g(1 - (2h)/(R))$g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states h ll R$h \ll R$. Equating the approximated form to depth gives hheight = (1)/(2) hdepth$h_{\text{height}} = \frac{1}{2} h_{\text{depth}}$, which fails when looking for a single unified distance value h$h$.
Chapter Mix
Class 11 Physics: Gravitation
Visual representation of points p and q showing positions above and below Earth's surface for Q47
More Gravitation Previous-Year Questions — Page 3
Q6jee_main_2025_07_april_eveningKepler's Laws of Planetary Motion
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant. [cite: 55]
Reason (R) For a central force field the angular momentum is a constant. [cite: 56]
In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 57]
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 63]
B. Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 65]
Kepler's Second Law state that the areal velocity (dA)/(dt)$\frac{dA}{dt}$ is directly proportional to the angular momentum L$L$ of the planet[cite: 55, 684]. Because the gravitational force between the Sun and the planet acts strictly along the line joining their centers (a central force field), its torque τ = r × F = 0$\vec{\tau} = \vec{r} \times \vec{F} = 0$[cite: 56, 686]. Since torque is zero, the angular momentum L$L$ remains completely constant over time[cite: 686]. As a consequence, (dA)/(dt) = constant$\frac{dA}{dt} = \text{constant}$[cite: 55, 684]. Both statements are true and (R) is the correct explanation of (A)[cite: 63].
Pattern Recognition
Areal velocity constancy is a direct geometric manifestation of the conservation of angular momentum under any central force field[cite: 55, 56, 686].
Chapter Mix
Class 11 Physics: Gravitation
Q25jee_main_2025_24_jan_eveningAcceleration due to Gravity
Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ____ g.
Numerical Answer.Answer: 9 to 9
Solution
Related Formula
Surface gravitational acceleration:
g = (GM)/(R²)$$g = \frac{GM}{R^2}$$
Core Logic
Since diameter drops to 1/3$1/3$, the radius R'$R'$ also scales down to 1/3$1/3$ its original value (R' = R/3$R' = R/3$), while mass M$M$ remains constant.
Gravity follows an inverse-square law with respect to radius. Shrinking the radius by a factor of n$n$ increases surface gravity by n²$n^2$ if the mass is unchanged.
Chapter Mix
Class 11 Physics: Gravitation
Qjee_main_2025_24_jan_morningKepler's Laws of Planetary Motion
A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-
A. 3%
B. 4.5%
C. 9%
D. 2.5%
Solution
Related Formula
By Kepler's Third Law of Planetary Motion, the square of the orbital period T$T$ is proportional to the cube of the orbital radius R$R$:
T² = K · R³$$T^{2} = K \cdot R^{3}$$
Core Logic
Taking logs and differentiating to find fractional errors for small changes:
Power factors act as direct linear multipliers for small percentage shifts. Here, the scaling factor is simply (3)/(2)$\frac{3}{2}$ times the radius change.
Chapter Mix
Class 11 Physics: Gravitation
Q11jee_main_2025_28_jan_eveningEscape Velocity
Earth has mass 8 \times and radius 2 \times that of a planet. If the escape velocity from the earth is 11.2 km/s$11.2 \, \text{km/s}$ , the escape velocity in km/s$\text{km/s}$ from the planet will be:
A. 11.2
B. 5.6
C. 2.8
D. 8.4
Solution
Related Formula
The expression for escape velocity from a spherical planetary body is given by:
Setting up quick ratios prevents substitution mistakes. For any planetary variant, notice how scaling properties scale inside the root operator directly.
Chapter Mix
Class 11 Physics: Gravitation
Q32jee_main_2024_01_february_morningAcceleration Due to Gravity
If R$R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π² ~m/s²$g = \pi^2 \mathrm{~m/s^2}$, then the length of the second's pendulum at a height h = 2R$h = 2R$ from the surface of earth will be:
1 = 9l l = (1)/(9)~m$$1 = 9l \implies l = \frac{1}{9}\mathrm{~m}$$
Pattern Recognition
Second's pendulum always has T = 2~s$T = 2\mathrm{~s}$. Height 2R$2R$ from the surface means a total distance of 3R$3R$ from the center, which yields a (1)/(9)$\frac{1}{9}$ drop in gravity.
Chapter Mix
Class 11 Physics: Gravitation
Class 11 Physics: Oscillations
More Gravitation Questions — jee_main_2024_29_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
