For a satellite orbiting close to the earth, orbital velocity is given by v0=g(R+h)=gR Escape velocity (ve) is ve=2g(R+h)=2gR[∵h<cR Δv=ve−v0=(2−1)gR
A satellite is revolving in a circular orbit at a height h from the carth surface, such that h<<R where R is the radius of the earth. Assuming that the effect of earth"s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is
Held on 11 Jan 2019 · Verified 6 Jul 2026.
2gR
gR
2gR
gR(2−1)
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Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
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Work through every JEE Main Mechanics PYQ, year by year.