The escape velocity of a bullet from the surface of planet Y is 1919.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.37 times that of Y, and its radius is 0.935 times the radius of Y

1584.9 m/s

Explanation:

Escape velocity on planet X is given as:

v = √(2*G*M/R)

Where G = gravitational constant

M = mass of planet X

R = radius of planet X

=> 1919 = √(2 * 6.67 * 10^(-11) * M/R)

1919² = (2 * 6.67 * 10^(-11)) * M/R

=> M/R = 3682561/(2 * 6.67 * 10^(-11))

M/R = 2.76 * 10^16 kg/m

Given that

Mass of planet X, M = 1.37 * mass of planet Y, m

m = M/1.37 = 0.73M

Radius of planet X, R = 0.935 * radius of planet Y, r

=> r = R/0.935 = 1.07R

Hence, the escape velocity on planet Y is:

v = √(2 * 6.67 * 10^(-11) * m/r)

Inputting the values of m and r in terms of M and R:

v = √(2 * 6.67 * 10^(-11) * 0.73M/1.07R)

Given that M/R = 2.76 * 10^16:

v = √(2 * 6.67 * 10^(-11) * 0.73 * 2.76 * 10^16 / 1.07)

v = √(25.12 * 10^5)

v = 1584.9 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west