Question

escape velocity of a body from earth is 11.2 km/s find the value of escape velocity of a body on the surface of Mars. Mars has a mass 1/9 and radius 1/2 that of the earth​

Answer 1

Given :

• Escape velocity of body from earth ($\sf V_{e_1}$ ) = 11.2 km/s
• Radius of Mars = (1/2) Radius of Earth
• Mass of Mars = (1/9) mass of Earth

Let :

• Mass of Earth = M
• Radius of Earth = R

To find :

Escape velocity of body from Mars

Formula used :

Escape velocity $\sf V_{e}$ = $\sf \sqrt{\dfrac{GM}{R} }$

Here

• M = Mass of planet
• R = Radius of planet
• G = Gravitational's constant

Solution :

✪ For earth

◕ Escape velocity ($\sf V_{e_1}$ )= $\sf \sqrt{\dfrac{GM}{R} }$

➝ 11.2 km/s = $\sf \sqrt{\dfrac{GM}{R} }$ ...... equation 1

✪ For Mars

◕ Radius = (1/2) of Earth

➝ Radius (R') = R ÷ 9

◕ Mass = (1/9) of Earth

➝ Mass (M') = M÷9

◕ Escape velocity ($\sf V_{e_2}$) = $\sf \sqrt{\dfrac{GM'}{R'} }$

➝ $\sf V_{e_2}$ = $\sf \sqrt{\dfrac{G(M/9)}{(R/2)} }$

➝ $\sf V_{e_2} = \sf \sqrt{\dfrac{GM}{R} }\times \sqrt{\dfrac{2}{9}}$

Putting value of , $\sf \sqrt{\dfrac{GM}{R} }$ from equation 1 , we get;

➝ $\sf V_{e_2} = \sf 11.2 \times \sqrt{\dfrac{2}{9}}$

➝ $\sf V_{e_2} = \sf 11.2 \times 0.47$

➝ $\sf V_{e_2} = 5.27\:km/s$

ANSWER :

Escape velocity of body from Mars = 5.27 km/h