Question

A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceleration at the surface of such a star?_______m/s2(b) How fast would an object be moving if it fell from rest through a distance of 16 m on such a star? (Assume the star does not rotate.)_______m/s

Answer 1

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s