MPSC Rajyaseva 2020 CSAT — Question 24

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पृथ्वीच्या केंद्रापासून R या अंतरावर एक उपग्रह गोलकार कक्षेत फिरत आहे. उपग्रह ϕ या कोनातून फिरताना व्यापले जाणारे क्षेत्र A = R²ϕ/2 आहे. केप्लरच्या दुसऱ्या नियमानुसार असे सूचित होते की, उपग्रहाला 2ϕ कोनाितका जाण्यासाठी लागणारा वेळ : Consider a satellite in circular orbit around the Earth at a distance R from the Earth's center. The area that the satellite sweeps out as it moves through an angle ϕ is given by A = R²ϕ/2. Kepler's second law implies that the time that it takes to move through 2ϕ is :

a)ϕ कोनातून जाण्याच्या दुप्पट वेळ लागेल. / double the time that it takes to move through ϕ.&check; Correct

b)ϕ कोनातून जाण्याच्या अर्धा वेळ लागेल. / half the time that it takes to move through ϕ.

c)ϕ कोनातून जाण्याच्या वेळेच्या वर्गाइतका वेळ लागेल. / the square of the time that it takes to move through ϕ.

d)ϕ कोनातून जाण्या इतकाच समान वेळ लागेल. / equal to the time that it takes to move through ϕ.

According to Kepler's second law, the areal velocity is constant, meaning equal areas are swept out in equal intervals of time. Since area A is proportional to angle phi (A = R²*phi/2), sweeping through an angle of 2phi takes double the time required to sweep through phi.