The ratio of escape velocity at the earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of the earth is (x/y √z). Find (x+y+z).

Question

The ratio of escape velocity at the earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of the earth is (x/y √z). Find (x+y+z). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Escape velocity =√(2 G M/R) =√(2 G/R) ⋅((4/3) ⋅ π R3 ρ) propto R √ρ Ratio, (ve/vp)=1: 2 √2

Q. The ratio of escape velocity at the earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of the earth is $\frac{x}{y z}$ . Find $(x + y + z)$ .

Escape velocity $= \frac{2 GM}{R}$
$= \frac{2 G}{R} \cdot (\frac{4}{3} \cdot π R^{3} ρ) \propto R ρ$
Ratio, $\frac{v _{e}}{v _{p}} = 1 : 22$

Q. The ratio of escape velocity at the earth (ve​) to the escape velocity at a planet (vp​) whose radius and mean density are twice as that of the earth is yz​x​. Find (x+y+z).

Escape velocity =R2GM​​ =R2G​⋅(34​⋅πR3ρ)​∝Rρ​ Ratio, vp​ve​​=1:22​

Q. The ratio of escape velocity at the earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of the earth is $\frac{x}{y z}$ . Find $(x + y + z)$ .

Escape velocity $= \frac{2 GM}{R}$
$= \frac{2 G}{R} \cdot (\frac{4}{3} \cdot π R^{3} ρ) \propto R ρ$
Ratio, $\frac{v _{e}}{v _{p}} = 1 : 22$