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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Solve the following problem.Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 1024 kg, R = 6400 km).

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प्रश्न

Solve the following problem.

Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 1024 kg, R = 6400 km).

बेरीज
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उत्तर

Given: h = 300 km = 0.3 × 106 m, M = 5.98 × 1024 kg, R = 6400 km = 6.4 × 106 m, G = 6.67 × 10-11 Nm2/kg

To find: Acceleration due to gravity at height (gh)

Formula: `"g"_"h" = "GM"/("R + h")^2`

Calculation:

From formula,

`"g"_"h" = (6.67 xx 10^-11 xx 5.98 xx 10^24)/([(6.4 xx 10^6) + (0.3 xx 10^6)]^2)`

`= (6.67 xx 5.98 xx 10^13)/((6.7)^2 xx 10^12)`

= antilog{log(6.67) + log(5.98) - 2log(6.7)} × 10

= antilog{0.8241 + 0.7767 - 2(0.8261)} × 10

= antilog{1.6008 - 1.6522} × 10

= antilog{`bar1`.9486} × 10

= 0.8884 × 10

= 8.884 m/s2

Acceleration due to gravity at 300 km will be 8.884 m/s2.

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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. (vi) | पृष्ठ ९९

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