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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Derive an expression for the critical velocity of a satellite.

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प्रश्न

Derive an expression for the critical velocity of a satellite.

व्युत्पत्ती
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उत्तर

The expression for critical velocity:

  1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
  2. If the satellite is moving in a circular orbit of the radius (R + h) = r, its speed must be equal to the magnitude of critical velocity vc.
  3. The centripetal force necessary for the circular motion of a satellite is provided by the gravitational force exerted by the satellite on the Earth.
    ∴ Centripetal force = Gravitational force
    ∴ `"mv"_"c"^2/"r" = "GMm"/"r"^2`
    ∴ `"v"_"c"^2 = "GM"/"r"`
    ∴ `"v"_"c" = sqrt("GM"/"r")`
    ∴ `"v"_"c" = sqrt("GM"/("R+h")) = sqrt("g"_"h" ("R + h"))`
    This is the expression for critical speed at the orbit of radius (R + h).
  4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is the revolving or gravitational acceleration at that altitude.
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पाठ 5: Gravitation - Exercises [पृष्ठ ९८]

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बालभारती Physics [English] Standard 11 Maharashtra State Board
पाठ 5 Gravitation
Exercises | Q 3. (i) | पृष्ठ ९८

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