The escape velocity from earth is 11 km-sec- The escape velocity from a planet having twice the radius and the same mean density as that of the earth is-1122 33sqrt - 3 - 44sqrt - 3 -

Question

Toppr Admin · Accepted Answer

Escape velocity ve-x221A-2GMeRe-xE001-xE000-xE000-xE000-x23B7-2G-43-x3C0-x3C1-R2ed-Re-x221A-2G-43-x3C0-x3C1-R2e-d-gt-ve-x3B1-Rsince density is same-when radius is doubled escape velocity gets doubled2ve-2-xD7-11-22kmS

The escape velocity from earth is 11 km/sec. The escape velocity from a planet having twice the radius and the same mean density as that of the earth is?
$11$
$22$
$33 \sqrt{3}$
$44 \sqrt{3}$

Escape velocity $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$

= $\sqrt{\frac{2 G (\frac{4}{3} π ρ R_{e}^{2} d)}{R_{e}}}$

$\sqrt{2 G (\frac{4}{3} π ρ R_{e}^{2}}) d => v_{e} α R$

since density is same.

when radius is doubled escape velocity gets doubled

$2 v_{e} = 2 \times 11 = 22 \frac{k m}{S}$

The escape velocity from earth is 11 km/sec. The escape velocity from a planet having twice the radius and the same mean density as that of the earth is?112233√344√3

Escape velocity ve=√2GMeRe=  ⎷2G(43πρR2ed)Re√2G(43πρR2e)d=>veαRsince density is same.when radius is doubled escape velocity gets doubled2ve=2×11=22kmS

The escape velocity from earth is 11 km/sec. The escape velocity from a planet having twice the radius and the same mean density as that of the earth is?
$11$
$22$
$33 \sqrt{3}$
$44 \sqrt{3}$

Escape velocity $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$

= $\sqrt{\frac{2 G (\frac{4}{3} π ρ R_{e}^{2} d)}{R_{e}}}$

$\sqrt{2 G (\frac{4}{3} π ρ R_{e}^{2}}) d => v_{e} α R$

since density is same.

when radius is doubled escape velocity gets doubled

$2 v_{e} = 2 \times 11 = 22 \frac{k m}{S}$