किसी उपग्रह के अधिकतम एवं न्यूनतम कक्षीय वेग क्रमशः vmax तथा vmin. हैं। सिद्ध करो कि उपग्रह की कक्षा की उत्केन्द्रता
प्रश्न : किसी उपग्रह के अधिकतम एवं न्यूनतम कक्षीय वेग क्रमशः vmax तथा vmin. हैं। सिद्ध करो कि उपग्रह की कक्षा की उत्केन्द्रता ε = Vmax – Vmin /Vmax + Vmin होती है।
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हल- ε उत्केन्द्रता वाले पथ का समीकरण है
R = p/1 + ε cos θ …………………..(1)
यहाँ p = J2/m K
p = J2/Mk
r + rε cos θ = p
t के सापेक्ष अवकलन करने पर
(dr/dt) + dr/dt εcos θ – rε sin θ dθ/dt = 0
Dr/dt = dθ /dt rε sin θ/1 + ε cos θ ……………………….(2)
उपग्रह का वेग v = [(dr/dt)2 + r2 (dθ/dt)2]1/2
Dr/dt का मान समीकरण (2) से रखने पर
V = r (dθ/dt) [1 + 2ε cos θ + ε2/(1 + εcos θ)2 ]
उपग्रह का कोणीय संवेग J = mr2 (dθ/dt)
V = J/mr [1 + 2ε cos θ + ε2/(1 + εcos θ)2 ]1/2 …………………..(3)
यदि θ = π हो तो v = vmin तथा r = rmax
vmin = J/mrmax तथा rmax = p/1 – ε
और यदि θ = 0 हो तो v = vmax तथा r = rmin
Vmax = J/mrmax = तथा rmax = p 1+ ε
Vmax = J(1 + ε)/mp
Vmax – Vmin/Vmax + Vmin = 1 + ε – 1 + ε/1 + ε + 1 – ε = ε
प्रश्न : निम्न केन्द्रीय बल के आधीन m द्रव्यमान का एक पिण्ड गतिमान है :
= – k r/r4
सिद्ध करो कि यदि पिण्ड की ऊर्जा E हो तो पिण्ड का वेग v = √k/mr2 + 2E/m
हल- केन्द्रीय बल
F (r) = K r/r4 = F(r) r = F (r) r/r
F (r) = – K/r3
अतः स्थितिज ऊर्जा U= – [F.dr = – K/2r2
कुल ऊर्जा E = 1/2 m (dr/dt)2 + J2/2mr2 – K/2r2
कोणीय संवेग J = mr2 (dθ/dt)
E = 1/2 m (dr/dt)2 + r2 (dθ/dt)2] – K/2r2
पिण्ड का वेग =(e) – (A)/ होता है।
पिण्ड का वेग v = [(dr/dt)2 + r2 (dθ/dt)2]1/2
E = 1/2 mv2 – k/2r2
या v = [K/mr2 + 2E/m]1/2
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Hindi sociology physics physical education maths english economics geography History
chemistry business studies biology accountancy political science
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Hindi physics physical education maths english economics
chemistry business studies biology accountancy Political science History sociology
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Class 6
Hindi social science science maths English
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Hindi social science science maths English
Class 8
Hindi social science science maths English
Class 9
Hindi social science science Maths English
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Hindi Social science science Maths English
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