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Earth revolves around the Sun at 30 km s−1. Calculate the kinetic energy of the Earth. In the previous example, you calculated the potential energy of the Earth. - Physics

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प्रश्न

Earth revolves around the Sun at 30 km s−1. Calculate the kinetic energy of the Earth. In the previous example, you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.

संख्यात्मक

उत्तर

Given: V = 30 km s−1

Potential energy = − 49.84 × 1032 J

Mass of earth = 5.9 × 1024

Formula:

Kinetic energy (K.E.) = `1/2"MV"^2`

= `1/2 xx 5.9 xx 10^24 xx 30 xx 30 xx 10^32`

= `1/2 xx 5.9 xx 3 xx 3 xx 10^32`

K.E. = 26.55 × 1032 J

∴ T.E. = K.E. – P.E.

T.E. = 26.55 × 1032 – 49.84 × 1032

T.E. = – 23.29 × 1032 J

Total energy is negative. It implies Earth is bounded with Sun.

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Gravitational Field and Gravitational Potential
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पाठ 6: Gravitation - Evaluation [पृष्ठ ४६]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 6 Gravitation
Evaluation | Q V. 12. | पृष्ठ ४६
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