Escape velocity earth-s surface is 11-2 km-s- Escape velocity the surface of a planet having mass 100 times and radius 4 times that of earth will be56 km-s112 km-s305 km-s11-2 km-s

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Toppr Admin · Accepted Answer

Ve-x221A-2gRVe-x221A-2GMR-x2234-g-GMR2-Now-xA0- -xA0- -xA0- -xA0-11-1Km-hr-x221A-2GMR-xA0- -xA0- -xA0- -xA0-V1e-x221A-2G-100M-4R-xA0- -xA0- -xA0- -xA0-11-1V1e-15-xA0- -xA0- -xA0- -xA0-V1e-55-5 Km-hr

Escape velocity at earth's surface is $11.2 k m / s$ . Escape velocity at the surface of a planet having mass $100$ times and radius $4$ times that of earth will be
$56 k m / s$
$112 k m / s$
$305 k m / s$
$11.2 k m / s$

$V_{e} = \sqrt{2 g R}$
$V_{e} = \sqrt{\frac{2 G M}{R}} [∴ g = \frac{G M}{R^{2}}]$
Now,
$11.1 K m / h r = \sqrt{\frac{2 G M}{R}}$
$V_{e}^{1} = \sqrt{\frac{2 G (100 M)}{4 R}}$
$\frac{11.1}{V_{e}^{1}} = \frac{1}{5}$
$V_{e}^{1} = 55.5$ Km/hr

Escape velocity at earth's surface is 11.2 km/s. Escape velocity at the surface of a planet having mass 100 times and radius 4 times that of earth will be56 km/s112 km/s305 km/s11.2 km/s

Ve=√2gRVe=√2GMR[∴g=GMR2]Now, 11.1Km/hr=√2GMR V1e=√2G(100M)4R 11.1V1e=15 V1e=55.5 Km/hr

Escape velocity at earth's surface is $11.2 k m / s$ . Escape velocity at the surface of a planet having mass $100$ times and radius $4$ times that of earth will be
$56 k m / s$
$112 k m / s$
$305 k m / s$
$11.2 k m / s$

$V_{e} = \sqrt{2 g R}$
$V_{e} = \sqrt{\frac{2 G M}{R}} [∴ g = \frac{G M}{R^{2}}]$
Now,
$11.1 K m / h r = \sqrt{\frac{2 G M}{R}}$
$V_{e}^{1} = \sqrt{\frac{2 G (100 M)}{4 R}}$
$\frac{11.1}{V_{e}^{1}} = \frac{1}{5}$
$V_{e}^{1} = 55.5$ Km/hr