Question

The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.59 times that of Y, and its radius is 0.903 times the radius of Y.

Answer 1

The escape velocity from the surface of the planet X is 2,249.2 m/s.

Escape velocity of planet X

$v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}$

where;

• M is mass of the planet
• r is radius of the planet
• G is universal gravitation constant

$\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s$

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

Learn more about escape velocity here: brainly.com/question/13726115

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