Gravitation
1. Kepler's Laws
Three laws governing planetary motion around the Sun.
1. Law of Orbits: Planets move in elliptical orbits with Sun at one focus.
2. Law of Areas: Areal velocity is constant (dA/dt = L/2m). (Conservation of Angular Momentum).
3. Law of Periods: T2 ∝ R3 (Square of time period is proportional to cube of semi-major axis).
2. Law of Areas: Areal velocity is constant (dA/dt = L/2m). (Conservation of Angular Momentum).
3. Law of Periods: T2 ∝ R3 (Square of time period is proportional to cube of semi-major axis).
(Asked in NEET 2018, 2021)
2. Universal Law of Gravitation
Every particle attracts every other particle with a force proportional to product of masses and inversely proportional to square of distance.
F = - (G m1 m2 / r2) r̂
G = 6.67 × 10-11 N m2/kg2.
G = 6.67 × 10-11 N m2/kg2.
Acceleration due to Gravity (g)
Surface: g = GM / R2
Height h: gh = g (1 - 2h/R) for h << R.
Depth d: gd = g (1 - d/R).
Center of Earth: g = 0.
Height h: gh = g (1 - 2h/R) for h << R.
Depth d: gd = g (1 - d/R).
Center of Earth: g = 0.
(Asked in NEET 2016, 2019, 2020, 2022)
3. Potential & Energy
Potential (V): -GM/r
Potential Energy (U): -GMm/r (Zero at infinity).
Work done W = ΔU = m(V2 - V1).
Potential Energy (U): -GMm/r (Zero at infinity).
Work done W = ΔU = m(V2 - V1).
Escape Velocity
Minimum speed to escape gravitational pull.
ve = √(2GM/R) = √(2gR)
For Earth, ve = 11.2 km/s.
For Earth, ve = 11.2 km/s.
4. Satellite Motion
Body revolving around a planet.
Orbital Speed (vo): √(GM/r) =
√(GM/(R+h)).
Near surface: vo = √(gR) ≈ 8 km/s.
Relation: ve = √2 vo.
Near surface: vo = √(gR) ≈ 8 km/s.
Relation: ve = √2 vo.
Energy of Satellite
KE = ½ m vo2 = GMm / 2r
PE = -GMm / r
Total Energy E = -GMm / 2r (Negative means bound system).
PE = -GMm / r
Total Energy E = -GMm / 2r (Negative means bound system).
Geostationary Satellite
- Time Period T = 24 Hours.
- Height h ≈ 36,000 km.
- Orbits in Equatorial Plane.
- Used for Communication.
(Asked in NEET 2017, 2019)
Numericals: Gravitation
Q1. Two spheres of mass 50 kg and 10 kg are 0.5 m apart. Calculate gravitational force between them.
Solution:
F = G m1 m2 / r2
F = (6.67 × 10-11 × 50 × 10) / (0.5)2
F = (6.67 × 10-11 × 500) / 0.25
F = 6.67 × 2000 × 10-11
F = 13340 × 10-11 = 1.33 × 10-7 N.
F = G m1 m2 / r2
F = (6.67 × 10-11 × 50 × 10) / (0.5)2
F = (6.67 × 10-11 × 500) / 0.25
F = 6.67 × 2000 × 10-11
F = 13340 × 10-11 = 1.33 × 10-7 N.
Q2. At what height above Earth's surface does the value of g become half of that on surface? (R = 6400 km)
Solution:
gh = g [R / (R+h)]2
g/2 = g [R / (R+h)]2
1/2 = [R / (R+h)]2 → 1/√2 = R / (R+h)
R+h = √2 R = 1.414 R
h = 0.414 R = 0.414 × 6400 ≈ 2650 km.
gh = g [R / (R+h)]2
g/2 = g [R / (R+h)]2
1/2 = [R / (R+h)]2 → 1/√2 = R / (R+h)
R+h = √2 R = 1.414 R
h = 0.414 R = 0.414 × 6400 ≈ 2650 km.
Q3. Find percentage decrease in weight of a body when taken to a depth of 32 km below surface.
Solution:
gd = g (1 - d/R).
Decrease = g - gd = g(d/R).
% Decrease = (Change/Original) × 100 = (d/R) × 100.
% Decrease = (32 / 6400) × 100 = 3200 / 6400 = 0.5 %.
gd = g (1 - d/R).
Decrease = g - gd = g(d/R).
% Decrease = (Change/Original) × 100 = (d/R) × 100.
% Decrease = (32 / 6400) × 100 = 3200 / 6400 = 0.5 %.
Q4. A planet has mass 4 times Earth and radius 2 times Earth. Find escape velocity if ve (Earth)
= 11.2 km/s.
Solution:
v = √(2GM/R).
v' = √(2G(4M)/(2R)) = √2 × √(2GM/R).
v' = √2 v = 1.414 × 11.2 ≈ 15.8 km/s.
v = √(2GM/R).
v' = √(2G(4M)/(2R)) = √2 × √(2GM/R).
v' = √2 v = 1.414 × 11.2 ≈ 15.8 km/s.
Q5. Calculate work done in lifting a body of mass 10 kg to a height equal to radius of Earth (R).
Solution:
W = ΔU = -GMm/(2R) - (-GMm/R)
W = GMm/2R = (gR2)m / 2R = mgR/2.
(Can use formula W = mgh/(1+h/R) with h=R → mgR/2).
W = 10 × 9.8 × 6.4×106 / 2
W = 98 × 3.2 × 106 = 3.13 × 108 J.
W = ΔU = -GMm/(2R) - (-GMm/R)
W = GMm/2R = (gR2)m / 2R = mgR/2.
(Can use formula W = mgh/(1+h/R) with h=R → mgR/2).
W = 10 × 9.8 × 6.4×106 / 2
W = 98 × 3.2 × 106 = 3.13 × 108 J.
Q6. A satellite revolves close to earth surface. Its KE is E. How much energy required to make it
escape?
Solution:
Orbital Velocity vo = √(GM/R). KE = ½mvo2 = GMm/2R = E.
Escape Velocity ve = √(2GM/R) = √2 vo.
Required KE for escape = ½mve2 = ½m(2vo2) = 2(½mvo2) = 2E.
Additional Energy Needed = 2E - E = E.
Orbital Velocity vo = √(GM/R). KE = ½mvo2 = GMm/2R = E.
Escape Velocity ve = √(2GM/R) = √2 vo.
Required KE for escape = ½mve2 = ½m(2vo2) = 2(½mvo2) = 2E.
Additional Energy Needed = 2E - E = E.
Q7. Distance between Earth and Moon is d. Mass of Earth is 81 times Moon. Where is gravitational
field zero?
Solution:
Field zero at distance x from moon.
G Mm / x2 = G ME / (d-x)2
1 / x2 = 81 / (d-x)2
Take square root: 1/x = 9 / (d-x)
d - x = 9x → d = 10x → x = d/10 from Moon.
Field zero at distance x from moon.
G Mm / x2 = G ME / (d-x)2
1 / x2 = 81 / (d-x)2
Take square root: 1/x = 9 / (d-x)
d - x = 9x → d = 10x → x = d/10 from Moon.
Q8. Geostationary satellite orbits at 6R from center. Find time period of another satellite at 3.5R.
Solution:
T2 ∝ R3.
(T2/T1)2 = (R2/R1)3
(T2/24)2 = (3.5R / 7R)3 (Note: h=6R means distance 7R usually, assuming questions means distance r). Let's assume distance given.
If r1=7R (approx geo), r2=3.5R = r1/2.
T22 = 242 (1/2)3 = 576 / 8 = 72.
T2 = √72 ≈ 8.5 hours.
T2 ∝ R3.
(T2/T1)2 = (R2/R1)3
(T2/24)2 = (3.5R / 7R)3 (Note: h=6R means distance 7R usually, assuming questions means distance r). Let's assume distance given.
If r1=7R (approx geo), r2=3.5R = r1/2.
T22 = 242 (1/2)3 = 576 / 8 = 72.
T2 = √72 ≈ 8.5 hours.
Q9. Three particles of mass m are at vertices of equilateral triangle of side a. Find PE of system.
Solution:
Pairs: (1,2), (2,3), (3,1). All distances same 'a'.
U = U12 + U23 + U31
U = -3 (Gmm / a) = -3 Gm2/a.
Pairs: (1,2), (2,3), (3,1). All distances same 'a'.
U = U12 + U23 + U31
U = -3 (Gmm / a) = -3 Gm2/a.
Q10. To what radius must Earth be compressed to become a Black hole? (M = 6×1024 kg)
Solution:
Schwarzschild Radius Rs = 2GM/c2
R = 2 × 6.67 × 10-11 × 6 × 1024 / (3 × 108)2
R = 80 × 1013 / 9 × 1016
R = 8.8 × 10-3 m ≈ 9 mm.
Schwarzschild Radius Rs = 2GM/c2
R = 2 × 6.67 × 10-11 × 6 × 1024 / (3 × 108)2
R = 80 × 1013 / 9 × 1016
R = 8.8 × 10-3 m ≈ 9 mm.
Important Formulae
20 NEET Golden Facts
- 1. Conservation of L: Kepler's 2nd Law (Law of Areas) is based on Conservation of Angular Momentum.
- 2. Center Force: Gravitational force is a Central Conservative force. Work done in closed path is zero.
- 3. Weightlessness: Inside a hollow spherical shell, field is zero, potential is constant (same as surface). Weight is zero.
- 4. Escape Speed: Independent of mass of projection body and angle of projection. Depends on Planet's mass and radius.
- 5. Maximum g: Value of g is maximum at Poles (Radius min) and minimum at Equator (Radius max + Rotation effect).
- 6. Binding Energy: Energy required to remove satellite from orbit to infinity. BE = +GMm/2r.
- 7. Geo-stationary: Must orbit in equatorial plane, West to East, T=24h, h=35800km.
- 8. Double Star: Two stars rotate about common center of mass with same angular speed ω.
- 9. No Atmosphere: Moon has no atmosphere because RMS speed of gas molecules > Escape velocity on Moon (2.3 km/s).
-
10. 1% Rule: For small % change in radius (h<
- 11. Inertial Mass: Measured by F/a. Gravitational Mass: Measured by Weight/g. They are equivalent.
- 12. Black Hole: Body whose escape velocity exceeds speed of light c. R = 2GM/c2 (Schwarzschild radius).
- 13. Tidal Waves: Caused by gravitational pull of Moon (and Sun) on Ocean water.
- 14. Pendulum: Time period of simple pendulum is infinite at center of earth (g=0).
- 15. Decrease in g: g decreases faster with height (2h) than with depth (d) near surface.
- 16. Speed at Infinity: If body thrown with v > ve, speed at infinity is √(v2 - ve2).
- 17. Work Done: To move mass m from surface to height h: W = mgh / (1 + h/R).
- 18. Orbital Speed: Decreases as radius increases (v ∝ 1/√r). Distant planets move slower.
- 19. Astronauts: Feel weightless in satellite because "Falling" freely towards earth (Normal reaction is zero).
- 20. Universal Constant: G is scalar and constant everywhere in universe. Independent of medium.
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