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Gravitation - Escape Speed and Orbital Velocity

Grade 11CBSEPhysics

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

🔑Concepts

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Escape Speed (vev_e): The minimum speed with which a body must be projected from the surface of a planet so that it escapes the gravitational pull of the planet and never returns. It is independent of the mass of the body and the angle of projection.

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Orbital Velocity (vov_o): The speed required to put a satellite into its orbit around a planet. For a satellite orbiting at a height hh above the Earth's surface, the centripetal force is provided by the gravitational force: mvo2(R+h)=GMm(R+h)2\frac{mv_o^2}{(R+h)} = \frac{GMm}{(R+h)^2}.

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Relationship between vev_e and vov_o: For a satellite orbiting very close to the Earth's surface (h≈0h \approx 0), the escape speed is 2\sqrt{2} times the orbital velocity, expressed as ve=2vov_e = \sqrt{2} v_o.

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Dependency on Mass and Radius: Both escape speed and orbital velocity are proportional to M\sqrt{M} and inversely proportional to R\sqrt{R}, where MM is the mass of the planet and RR is its radius.

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Atmospheric constraints: A planet (like the Moon) lacks an atmosphere if the root mean square (RMS) speed of its gas molecules is greater than its escape speed, causing the gas to escape into space.

📐Formulae

ve=2GMRv_e = \sqrt{\frac{2GM}{R}}

ve=2gRv_e = \sqrt{2gR}

vo=GMR+hv_o = \sqrt{\frac{GM}{R+h}}

vo=gR (for h≪R)v_o = \sqrt{gR} \text{ (for } h \ll R\text{)}

ve=2vov_e = \sqrt{2} v_o

💡Examples

Problem 1:

Determine the escape velocity on the surface of Mars. Given that the mass of Mars is 6.42×1023 kg6.42 \times 10^{23} \text{ kg} and its radius is 3.39×106 m3.39 \times 10^6 \text{ m}. Take G=6.67×10−11 N m2kg−2G = 6.67 \times 10^{-11} \text{ N m}^2\text{kg}^{-2}.

Solution:

Using the formula ve=2GMRv_e = \sqrt{\frac{2GM}{R}}: ve=2×6.67×10−11×6.42×10233.39×106≈5027 m/s≈5.03 km/sv_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.39 \times 10^6}} \approx 5027 \text{ m/s} \approx 5.03 \text{ km/s}.

Explanation:

The escape velocity is calculated by substituting the universal gravitational constant, the mass of Mars, and the radius of Mars into the escape speed formula derived from the principle of conservation of energy.

Problem 2:

A satellite is orbiting close to the Earth's surface. If the orbital velocity is 7.92 km/s7.92 \text{ km/s}, calculate the escape velocity for the Earth.

Solution:

ve=2×vo=1.414×7.92 km/s≈11.2 km/sv_e = \sqrt{2} \times v_o = 1.414 \times 7.92 \text{ km/s} \approx 11.2 \text{ km/s}.

Explanation:

For orbits near the surface, the relation ve=2vov_e = \sqrt{2}v_o holds. Multiplying the given orbital velocity by 2\sqrt{2} provides the escape velocity of 11.2 km/s11.2 \text{ km/s}.

Escape Speed and Orbital Velocity - Revision Notes & Key Formulas | CBSE Class 11 Physics