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Gravitation - Earth Satellites

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A satellite is any body that revolves around a planet in a stable orbit under the influence of the planet's gravitational pull.

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The centripetal force required for the circular motion of a satellite is provided by the gravitational force between the Earth and the satellite: Fc=FgF_c = F_g.

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Orbital Velocity (vov_o) is the specific velocity required for a satellite to remain in a stable circular orbit at a height hh above the Earth's surface.

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The Time Period (TT) of a satellite is the time it takes to complete one full revolution around the Earth. It depends on the radius of the orbit: T2∝r3T^2 \propto r^3 (Kepler's Third Law).

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Geostationary Satellites revolve in the equatorial plane from West to East with a period of T=24T = 24 hours. They appear stationary relative to an observer on Earth and are used for telecommunications.

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Polar Satellites revolve in orbits that pass over the North and South poles. They are much closer to Earth (hβ‰ˆ500h \approx 500 to 800800 km) and are used for remote sensing and meteorology.

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Weightlessness is a state where the effective weight of an object becomes zero. Inside a satellite, the acceleration of the satellite equals the acceleration due to gravity, leading to a normal reaction force of zero: N=m(gβˆ’a)=0N = m(g - a) = 0.

πŸ“Formulae

vo=GMeRe+h=RegRe+hv_o = \sqrt{\frac{GM_e}{R_e + h}} = R_e\sqrt{\frac{g}{R_e + h}}

T=2Ο€(Re+h)3GMe=2Ο€Re(Re+h)3gT = 2\pi \sqrt{\frac{(R_e + h)^3}{GM_e}} = \frac{2\pi}{R_e} \sqrt{\frac{(R_e + h)^3}{g}}

K.E.=GMem2(Re+h)K.E. = \frac{GM_em}{2(R_e + h)}

P.E.=βˆ’GMemRe+hP.E. = -\frac{GM_em}{R_e + h}

Etotal=K.E.+P.E.=βˆ’GMem2(Re+h)E_{total} = K.E. + P.E. = -\frac{GM_em}{2(R_e + h)}

ve=2voΒ (forΒ orbitΒ nearΒ Earth’sΒ surface)v_e = \sqrt{2} v_o \text{ (for orbit near Earth's surface)}

πŸ’‘Examples

Problem 1:

Calculate the orbital velocity of a satellite orbiting very close to the Earth's surface. Given g=9.8Β m/s2g = 9.8 \text{ m/s}^2 and Re=6.4Γ—106Β mR_e = 6.4 \times 10^6 \text{ m}.

Solution:

For a satellite near the surface, hβ‰ˆ0h \approx 0. The formula for orbital velocity becomes: vo=gRev_o = \sqrt{gR_e} Substituting the values: vo=9.8Γ—6.4Γ—106v_o = \sqrt{9.8 \times 6.4 \times 10^6} vo=62.72Γ—106β‰ˆ7.92Γ—103Β m/sv_o = \sqrt{62.72 \times 10^6} \approx 7.92 \times 10^3 \text{ m/s} voβ‰ˆ7.92Β km/sv_o \approx 7.92 \text{ km/s}

Explanation:

When a satellite is close to the Earth's surface, the orbital radius is approximately equal to the Earth's radius (ReR_e). The velocity required to maintain this orbit is roughly 7.9Β km/s7.9 \text{ km/s}.

Problem 2:

A satellite of mass mm is moved from an orbit of radius 2Re2R_e to 3Re3R_e. Calculate the change in its total energy.

Solution:

The total energy of a satellite in orbit of radius rr is: E=βˆ’GMem2rE = -\frac{GM_em}{2r} Initial energy at r1=2Rer_1 = 2R_e: E1=βˆ’GMem4ReE_1 = -\frac{GM_em}{4R_e} Final energy at r2=3Rer_2 = 3R_e: E2=βˆ’GMem6ReE_2 = -\frac{GM_em}{6R_e} Change in energy Ξ”E=E2βˆ’E1\Delta E = E_2 - E_1: Ξ”E=βˆ’GMem6Reβˆ’(βˆ’GMem4Re)\Delta E = -\frac{GM_em}{6R_e} - \left( -\frac{GM_em}{4R_e} \right) Ξ”E=GMemRe(14βˆ’16)\Delta E = \frac{GM_em}{R_e} \left( \frac{1}{4} - \frac{1}{6} \right) Ξ”E=GMem12Re\Delta E = \frac{GM_em}{12R_e}

Explanation:

Energy must be supplied to move a satellite to a higher orbit. The change in energy is positive because the final total energy is less negative (higher) than the initial energy.