Question

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

Answer 1

Answer:

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Answer 2

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