A satellite is revolving around the earth in a circular orbit at a height R above the surface of the earth, where R is the radius of the earth. If the satellite is moved to a new circular orbit such that its time period of revolution becomes 8 times its initial time period, the new height of the satellite above the surface of the earth will be:

Step-by-step solution

Let R be the radius of the earth. The initial orbital radius of the satellite from the center of the earth is: r₁ = R + h₁ = R + R = 2R According to Kepler's third law, T^2 r^3 . ( r₂ r₁ )^3 = ( T₂ T₁ )^2 Given that the new time period T₂ = 8 T₁ , we have: ( r₂ r₁ )^3 = (8)^2 = 64 r₂ r₁ = (64)^ 1/3 = 4 r₂ = 4 r₁ = 4(2R) = 8R The new orbital radius from the center of the earth is 8R . The new height h₂ above the surface of the earth is: h₂ = r₂ - R = 8R - R = 7R Answer: 7R