Math, asked by Brainlyaccount, 1 year ago

एक उपग्रह पृथ्वी तल से 3400km की दूरी पर वृत्तीय कक्षा में घूम रहा है उपग्रह के कक्षीय वेग तथा परिक्रमण काल की गणना करो
पृथ्वी की तृज्या 6400km तथा गुरूत्वीय त्वरण g = 9.8m/s²


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Swarnimkumar22: ..
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Answers

Answered by Swarnimkumar22
24
Thanks for asking question



दिया है-

पृथ्वी की तृज्या 6400km तथा गुरूत्वीय त्वरण g = 9.8m/s²

तब,

Re + h => 6400+3400 = 9800km

 = 9.8 \times 10 {}^{6}

re \:  = 6.4 \times  {10}^{6}

कक्षीय वेग

 vo = re \sqrt{ \frac{g}{re + h} }


 = 6.4 \times 10 {}^{6}  \sqrt{ \frac{9.8}{9.8 \times 10 {}^{6} } }


 =  \frac{6.4 \times 10 {}^{6} }{ {10}^{3} }


6.4 \times  {10}^{3} km.s

t \:  =  \frac{2\pi(re + h)}{vo}


 \frac{2 \times 3.14 \times 9.8 \times  {10}^{6} }{6.4 \times  {10}^{3} }


 \frac{6.28 \times 9.8 \times  {10}^{3} }{6.4}


 \frac{61.544 \times  {10}^{3} }{6.4}


 = 9.7 \times  {10}^{3} sec



I hope it's help you
Answered by tanmaypandey2000
1

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