SATHEE: Gravitation 4 Question 5

Gravitation 4 Question 5

6. The ratio of earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$ . The area enclosed by earth’s orbit is approximately $m^{2}$ .

(1997C, 1M)

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Answer:

Correct Answer: 6. $6.94 \times 10^{22}$

Solution:

Areal velocity of a planet round the sun is constant and is given by

$\frac{d A}{d t} = \frac{L}{2 m}$

where, $L =$ angular momentum of planet (earth) about sun and $m =$ mass of planet (earth).

Given,

$\frac{L}{m} = 4.4 \times 10^{15} m^{2} / s$

$∴$ Area enclosed by earth in time $T$ ( 365 days ) will be

$\begin{aligned} Area & = \frac{d A}{d t} \cdot T = \frac{L}{2 m} \cdot T \\ = \frac{1}{2} \times 4.4 \times 10^{15} \times 365 \times 24 \times 3600 m^{2} \end{aligned}$

$Area \approx 6.94 \times 10^{22} m^{2}$

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Gravitation 4 Question 5

6. The ratio of earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$ . The area enclosed by earth’s orbit is approximately $m^{2}$ .

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Gravitation 4 Question 5

6. The ratio of earth’s orbital angular momentum (about the sun) to its mass is 4.4×1015m2/s. The area enclosed by earth’s orbit is approximately m2.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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6. The ratio of earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$ . The area enclosed by earth’s orbit is approximately $m^{2}$ .