Physics, asked by Humaira5970, 8 months ago

A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. then the time period of a spy satellite orbiting a frw hundred km (600 km) above the earth's surface (R=6400 km) will approximately be

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Answered by Anonymous

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Answered by Theopekaaleader

$Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1Solve the following by both substitution and elimination methods.2x - \sqrt{2} y = 02x− 2 y=0and\frac{3x}{ \sqrt{2} } - y = 1 2 3x −y=1$

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