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Gravitation - Orbital Velocity of a Satellite

Grade 11ICSEPhysics

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

🔑Concepts

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Orbital velocity (vov_o) is the minimum velocity required to put a satellite into a stable circular orbit around a celestial body.

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The necessary centripetal force for the circular motion is provided by the gravitational force of attraction between the planet and the satellite: mvo2r=GMmr2\frac{mv_o^2}{r} = \frac{GMm}{r^2}.

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Orbital velocity is independent of the mass of the satellite (mm). It depends only on the mass of the central body (MM) and the radius of the orbit (rr).

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The radius of the orbit is r=R+hr = R + h, where RR is the radius of the planet and hh is the height of the satellite above the surface.

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If a satellite orbits very close to the Earth's surface (h≈0h \approx 0), the orbital velocity is approximately 7.92 km/s7.92 \text{ km/s}.

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The relationship between escape velocity (vev_e) and orbital velocity near the Earth's surface is given by ve=2vov_e = \sqrt{2} v_o.

📐Formulae

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

vo=gR2R+hv_o = \sqrt{\frac{gR^2}{R+h}}

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

T=2π(R+h)vo=2π(R+h)3GMT = \frac{2\pi(R+h)}{v_o} = 2\pi \sqrt{\frac{(R+h)^3}{GM}}

💡Examples

Problem 1:

Calculate the orbital velocity of a satellite revolving around the Earth at a height of 3600 km3600 \text{ km} above the surface. Given R=6400 kmR = 6400 \text{ km} and g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution:

  1. Calculate orbital radius: r=R+h=6400 km+3600 km=10000 km=107 mr = R + h = 6400 \text{ km} + 3600 \text{ km} = 10000 \text{ km} = 10^7 \text{ m}.
  2. Use the formula: vo=gR2rv_o = \sqrt{\frac{gR^2}{r}}.
  3. Substitute values: vo=9.8×(6.4×106)2107v_o = \sqrt{\frac{9.8 \times (6.4 \times 10^6)^2}{10^7}}.
  4. vo=9.8×40.96×1012107=401.408×105≈6335 m/sv_o = \sqrt{\frac{9.8 \times 40.96 \times 10^{12}}{10^7}} = \sqrt{401.408 \times 10^5} \approx 6335 \text{ m/s}.

Explanation:

The orbital velocity is found by relating the acceleration due to gravity at the surface to the gravitational pull at height hh. By substituting the total distance from the center of the Earth into the formula vo=Rg/rv_o = R\sqrt{g/r}, we find the required velocity.

Problem 2:

A satellite is orbiting very close to the surface of a planet with a mass 4 times that of Earth and a radius 2 times that of Earth. Find its orbital velocity in terms of Earth's orbital velocity vev_e.

Solution:

  1. Let Earth's orbital velocity be ve=GMRv_e = \sqrt{\frac{GM}{R}}.
  2. For the new planet: M′=4MM' = 4M and R′=2RR' = 2R.
  3. vo′=GM′R′=G(4M)2Rv_o' = \sqrt{\frac{GM'}{R'}} = \sqrt{\frac{G(4M)}{2R}}.
  4. vo′=2×GMR=2vev_o' = \sqrt{2} \times \sqrt{\frac{GM}{R}} = \sqrt{2} v_e.

Explanation:

Since orbital velocity is proportional to M/R\sqrt{M/R}, substituting the ratios of mass and radius allows us to find the scale factor relative to Earth's orbital velocity.

Orbital Velocity of a Satellite - Revision Notes & Key Formulas | ICSE Class 11 Physics