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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Solve the following problem. Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface.(ME = 5.98 × 1024 kg, R = 6.4 × 106 m)

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प्रश्न

Solve the following problem.

Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface.
(ME = 5.98 × 1024 kg, R = 6.4 × 106 m)

बेरीज
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उत्तर

Given: h = 1000 km = 1 × 106 m, ME = 5.98 × 1024 kg, R = 6.4 × 106 m, G = 6.67 × 10–11 N m2/kg2

To find: Speed of satellite (vc)

Formula: `"v"_"c" = sqrt("GM"/"r")`

Calculation: From formula,

`"v"_"c" = sqrt("GM"/(("R + h")))`

`= sqrt((6.67 xx 10^-11 xx 5.98 xx 10^24)/([(6.4 xx 10^6) + (1 xx 10^6)]))`

`= sqrt((6.67 xx 5.98 xx 10^7)/7.4)`

`= sqrt("antilog"{log (6.67) + log(5.98) - log(7.4) xx 10^7})`

`= sqrt("antilog"{0.8241 + 0.7767 - 0.8692} xx 10^7)`

`= sqrt("antilog"{0.7316} xx 10^7)`

`= sqrt(5.391 xx 10^7)`

`= sqrt(53.91 xx 10^6)`

= 7.343 × 103     .....[Taking square root value]

= 7.343 × 103 m/s

Speed of the satellite at height 1000 km is 7.343 × 103 m/s.

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पाठ 5: Gravitation - Exercises [पृष्ठ ९९]

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बालभारती Physics [English] Standard 11 Maharashtra State Board
पाठ 5 Gravitation
Exercises | Q 4. (vii) | पृष्ठ ९९

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