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Physics (SSC, Railway, Police & All State exam)Chapter Unit

Gravitation

Introduction to Gravitation

Gravitation is the force of attraction between any two masses in the universe. It is one of the fundamental forces of nature and governs phenomena ranging from the motion of planets to the falling of objects on Earth.


Newton's Law of Gravitation

  1. Statement:

    • Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
  2. Mathematical Expression: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}F=Gr2m1​m2​​

    • FFF: Gravitational force.
    • GGG: Gravitational constant (6.67×10−11 N m2 kg−26.67 \times 10^{-11} \, N \, m^2 \, kg^{-2}6.67×10−11Nm2kg−2).
    • m1m_1m1​, m2m_2m2​: Masses of the two objects.
    • rrr: Distance between the centers of the masses.
  3. Key Features:

    • It is a central force (acts along the line joining the two masses).
    • It is an attractive force.
    • It is universal and acts between all objects with mass.

Gravitational Field

  1. Definition:

    • The region around a mass where another mass experiences a gravitational force.
  2. Gravitational Field Intensity (ggg):

    • Force per unit mass experienced by a small test mass placed in the field: g=Fm=GMr2g = \frac{F}{m} = \frac{GM}{r^2}g=mF​=r2GM​
    • SI Unit: m/s2m/s^2m/s2 or N/kgN/kgN/kg.
  3. Variation of ggg with Altitude:

    • At a height hhh above the Earth's surface: gh=g(RR+h)2g_h = g \left( \frac{R}{R + h} \right)^2gh​=g(R+hR​)2
      • RRR: Radius of the Earth.

Acceleration Due to Gravity

  1. On the Surface of the Earth:

    • Formula: g=GMR2g = \frac{GM}{R^2}g=R2GM​
      • MMM: Mass of the Earth.
      • RRR: Radius of the Earth.
  2. Factors Affecting ggg:

    • Height (hhh): Decreases with altitude.
    • Depth (ddd): Decreases with depth: gd=g(1−dR)g_d = g \left( 1 - \frac{d}{R} \right)gd​=g(1−Rd​)
    • Latitude: Slightly less at the equator than at the poles due to Earth's rotation.

Gravitational Potential Energy

  1. Definition:

    • The energy possessed by a body due to its position in a gravitational field.
    • Formula: U=−GMmrU = -\frac{GMm}{r}U=−rGMm​
      • Negative sign indicates that gravitational potential energy decreases as the distance from the mass decreases.
  2. Work Done in Moving a Body: W=Ufinal−UinitialW = U_{\text{final}} - U_{\text{initial}}W=Ufinal​−Uinitial​


Escape Velocity

  1. Definition:

    • The minimum velocity required for an object to escape the gravitational pull of a planet without returning.
  2. Formula: ve=2GMR=2gRv_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}ve​=R2GM​​=2gR​

    • For Earth: ve≈11.2 km/sv_e \approx 11.2 \, km/sve​≈11.2km/s
  3. Key Points:

    • Escape velocity is independent of the mass of the object.
    • It depends only on the planet's mass and radius.

Numerical Example

  1. Example 1: Calculate the gravitational force between two masses, m1=10 kgm_1 = 10 \, kgm1​=10kg and m2=20 kgm_2 = 20 \, kgm2​=20kg, separated by a distance of 2 m2 \, m2m.

    • Formula: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}F=Gr2m1​m2​​
    • Substituting values: F=6.67×10−11⋅10⋅2022F = 6.67 \times 10^{-11} \cdot \frac{10 \cdot 20}{2^2}F=6.67×10−11⋅2210⋅20​ F=6.67×10−11⋅2004=3.34×10−9 NF = 6.67 \times 10^{-11} \cdot \frac{200}{4} = 3.34 \times 10^{-9} \, NF=6.67×10−11⋅4200​=3.34×10−9N
  2. Example 2: Calculate the escape velocity for a planet with M=6×1024 kgM = 6 \times 10^{24} \, kgM=6×1024kg and R=6.4×106 mR = 6.4 \times 10^6 \, mR=6.4×106m.

    • Formula: ve=2GMRv_e = \sqrt{\frac{2GM}{R}}ve​=R2GM​​
    • Substituting values: ve=2⋅6.67×10−11⋅6×10246.4×106v_e = \sqrt{\frac{2 \cdot 6.67 \times 10^{-11} \cdot 6 \times 10^{24}}{6.4 \times 10^6}}ve​=6.4×1062⋅6.67×10−11⋅6×1024​​ ve=8.004×10146.4×106=1.25×108≈11.2 km/sv_e = \sqrt{\frac{8.004 \times 10^{14}}{6.4 \times 10^6}} = \sqrt{1.25 \times 10^8} \approx 11.2 \, km/sve​=6.4×1068.004×1014​​=1.25×108​≈11.2km/s

Orbital Motion of Satellites

  1. Orbital Velocity:

    • The minimum velocity required to keep a satellite in a stable circular orbit around a planet.

    • Formula: vo=GMrv_o = \sqrt{\frac{GM}{r}}vo​=rGM​​

      • MMM: Mass of the planet.
      • rrr: Radius of orbit (sum of the planet's radius and the satellite's altitude).
    • For Earth: vo≈7.9 km/sv_o \approx 7.9 \, km/svo​≈7.9km/s

  2. Time Period of a Satellite:

    • The time taken by a satellite to complete one revolution.

    • Formula: T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}T=2πGMr3​​

      • rrr: Radius of orbit.
      • GGG: Gravitational constant.
      • MMM: Mass of the planet.
    • For a satellite near Earth: T≈90 minutesT \approx 90 \, \text{minutes}T≈90minutes

  3. Geostationary Satellites:

    • Satellites that appear stationary with respect to a point on Earth.
    • Characteristics:
      • Orbit at an altitude of approximately 36,000 km36,000 \, km36,000km.
      • Orbital period matches Earth’s rotation period (24 hours).
      • Used for communication and weather forecasting.

Kepler’s Laws of Planetary Motion

  1. First Law (Law of Ellipses):

    • Planets move in elliptical orbits with the Sun at one of the foci.
  2. Second Law (Law of Equal Areas):

    • A line joining a planet and the Sun sweeps out equal areas in equal intervals of time.
    • Implication:
      • Planets move faster when closer to the Sun (perihelion) and slower when farther (aphelion).
  3. Third Law (Law of Periods):

    • The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
    • Formula: T2∝r3orT2r3=constantT^2 \propto r^3 \quad \text{or} \quad \frac{T^2}{r^3} = \text{constant}T2∝r3orr3T2​=constant

Gravitational Potential

  1. Definition:

    • The gravitational potential (VVV) at a point is the work done per unit mass to bring a test mass from infinity to that point.
    • Formula: V=−GMrV = -\frac{GM}{r}V=−rGM​
      • MMM: Mass of the attracting body.
      • rrr: Distance from the center of the attracting body.
  2. Gravitational Potential Difference:

    • Work done to move a mass mmm between two points: W=m⋅ΔV=m(V2−V1)W = m \cdot \Delta V = m \left( V_2 - V_1 \right)W=m⋅ΔV=m(V2​−V1​)

Weightlessness

  1. Definition:

    • A condition where a body experiences no net gravitational force or normal force.
    • Occurs in free-fall situations or inside orbiting satellites.
  2. Explanation:

    • Inside a satellite, the gravitational force provides the centripetal force, and the normal reaction force becomes zero, creating a sensation of weightlessness.

Energy in Orbital Motion

  1. Kinetic Energy of a Satellite:

    • Formula: KE=12mvo2=GMm2rKE = \frac{1}{2} mv_o^2 = \frac{GMm}{2r}KE=21​mvo2​=2rGMm​
  2. Potential Energy of a Satellite:

    • Formula: PE=−GMmrPE = -\frac{GMm}{r}PE=−rGMm​
  3. Total Energy of a Satellite:

    • Total energy is the sum of kinetic and potential energy: TE=KE+PE=−GMm2rTE = KE + PE = -\frac{GMm}{2r}TE=KE+PE=−2rGMm​
    • The negative sign indicates the satellite is bound to the planet.

Escape and Orbital Velocities Relation

  1. Relation:

    • Escape velocity (vev_eve​) is related to orbital velocity (vov_ovo​) as: ve=2vov_e = \sqrt{2} v_ove​=2​vo​
  2. Implication:

    • Escape velocity is always greater than the orbital velocity for the same altitude.

Numerical Examples

  1. Example 1: Calculate the orbital velocity of a satellite orbiting the Earth at a height of 500 km500 \, km500km. (Take M=6×1024 kgM = 6 \times 10^{24} \, kgM=6×1024kg, R=6.4×106 mR = 6.4 \times 10^6 \, mR=6.4×106m).

    • Formula: vo=GMR+hv_o = \sqrt{\frac{GM}{R + h}}vo​=R+hGM​​
    • Substituting values: vo=6.67×10−11⋅6×10246.4×106+500×103v_o = \sqrt{\frac{6.67 \times 10^{-11} \cdot 6 \times 10^{24}}{6.4 \times 10^6 + 500 \times 10^3}}vo​=6.4×106+500×1036.67×10−11⋅6×1024​​ vo=4.002×10146.9×106≈7.6 km/sv_o = \sqrt{\frac{4.002 \times 10^{14}}{6.9 \times 10^6}} \approx 7.6 \, km/svo​=6.9×1064.002×1014​​≈7.6km/s
  2. Example 2: A geostationary satellite has an orbital radius of 42,000 km42,000 \, km42,000km. Calculate its time period and orbital velocity.

    • Time Period: T=24 hours=86,400 sT = 24 \, \text{hours} = 86,400 \, sT=24hours=86,400s
    • Orbital Velocity: vo=2πrT=2π⋅42,000×10386,400≈3.07 km/sv_o = \frac{2\pi r}{T} = \frac{2\pi \cdot 42,000 \times 10^3}{86,400} \approx 3.07 \, km/svo​=T2πr​=86,4002π⋅42,000×103​≈3.07km/s

Variation of Acceleration Due to Gravity (ggg)

  1. Variation with Altitude (hhh):

    • The value of ggg decreases with increasing altitude.
    • Formula: gh=g(RR+h)2g_h = g \left( \frac{R}{R + h} \right)^2gh​=g(R+hR​)2
      • RRR: Radius of the Earth.
      • hhh: Height above the Earth’s surface.
  2. Variation with Depth (ddd):

    • The value of ggg decreases as we move below the Earth’s surface.
    • Formula: gd=g(1−dR)g_d = g \left( 1 - \frac{d}{R} \right)gd​=g(1−Rd​)
      • ddd: Depth below the surface.
  3. Variation with Latitude:

    • Due to Earth's rotation, ggg is slightly less at the equator than at the poles.
    • Formula: geffective=g−Rω2cos⁡2ϕg_{\text{effective}} = g - R\omega^2 \cos^2 \phigeffective​=g−Rω2cos2ϕ
      • ω\omegaω: Angular velocity of Earth.
      • ϕ\phiϕ: Latitude.

Tides

  1. Cause:

    • Tides are caused by the gravitational pull of the Moon and the Sun on Earth's oceans.
  2. Types of Tides:

    • Spring Tides: Higher than usual tides when the Sun, Moon, and Earth are aligned (new moon and full moon phases).
    • Neap Tides: Lower than usual tides when the Sun and Moon are at right angles relative to the Earth (first and third quarter phases).

Gravitational Force vs. Electrostatic Force

PropertyGravitational ForceElectrostatic Force
NatureAlways attractiveCan be attractive or repulsive
Proportional toProduct of massesProduct of charges
Governing ConstantGravitational constant (GGG)Coulomb's constant (kkk)
Relative StrengthWeakStrong
RangeInfiniteInfinite

Artificial Satellites

  1. Types of Satellites:

    • Geostationary Satellites:
      • Appear stationary with respect to a point on Earth.
      • Used for communication and weather monitoring.
    • Polar Satellites:
      • Orbit the Earth in a north-south direction.
      • Used for mapping, environmental monitoring, and reconnaissance.
  2. Uses of Satellites:

    • Weather forecasting.
    • Global positioning systems (GPS).
    • Scientific research and exploration.

Black Holes

  1. Definition:

    • A black hole is a region in space where the gravitational pull is so strong that nothing, not even light, can escape.
  2. Schwarzschild Radius (rsr_srs​):

    • The radius of the event horizon of a black hole.
    • Formula: rs=2GMc2r_s = \frac{2GM}{c^2}rs​=c22GM​
      • MMM: Mass of the black hole.
      • ccc: Speed of light.

Numerical Examples

  1. Example 1: Calculate the variation in ggg at a height of 500 km500 \, km500km above the Earth’s surface. Take g=9.8 m/s2g = 9.8 \, m/s^2g=9.8m/s2 and R=6.4×106 mR = 6.4 \times 10^6 \, mR=6.4×106m.

    • Formula: gh=g(RR+h)2g_h = g \left( \frac{R}{R + h} \right)^2gh​=g(R+hR​)2
    • Substituting values: gh=9.8(6.4×1066.4×106+500×103)2g_h = 9.8 \left( \frac{6.4 \times 10^6}{6.4 \times 10^6 + 500 \times 10^3} \right)^2gh​=9.8(6.4×106+500×1036.4×106​)2 gh=9.8(6.46.9)2g_h = 9.8 \left( \frac{6.4}{6.9} \right)^2gh​=9.8(6.96.4​)2 gh=9.8⋅(0.927)2=9.8⋅0.859=8.42 m/s2g_h = 9.8 \cdot \left( 0.927 \right)^2 = 9.8 \cdot 0.859 = 8.42 \, m/s^2gh​=9.8⋅(0.927)2=9.8⋅0.859=8.42m/s2
  2. Example 2: Calculate the Schwarzschild radius of a black hole with a mass 10 M⊙10 \, M_{\odot}10M⊙​, where M⊙=2×1030 kgM_{\odot} = 2 \times 10^{30} \, kgM⊙​=2×1030kg.

    • Formula: rs=2GMc2r_s = \frac{2GM}{c^2}rs​=c22GM​
    • Substituting values: rs=2⋅6.67×10−11⋅10⋅2×1030(3×108)2r_s = \frac{2 \cdot 6.67 \times 10^{-11} \cdot 10 \cdot 2 \times 10^{30}}{(3 \times 10^8)^2}rs​=(3×108)22⋅6.67×10−11⋅10⋅2×1030​ rs=2.67×10219×1016≈29.7 kmr_s = \frac{2.67 \times 10^{21}}{9 \times 10^{16}} \approx 29.7 \, kmrs​=9×10162.67×1021​≈29.7km

Recap: Key Points to Remember

  • Newton’s law of gravitation governs the attractive force between two masses.
  • Acceleration due to gravity varies with altitude, depth, and latitude.
  • Satellites rely on gravitational forces for stable orbits.
  • Conservation laws and gravitational concepts apply to large-scale phenomena, including black holes and tides.

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