Question

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.017 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answer 1

Answer:

30298514.82 m/s

Explanation:

M = Mass of star = 2×10³ kg

r = Radius of star = 5×10³ m

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

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$a=G\frac{M}{r^2}\\\Rightarrow a=6.67\times 10^{-11}\frac{2\times 10^{30}}{5\times 10^3}\\\Rightarrow a=2.7\times 10^{16}\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 2.7\times 10^{16}\times 0.017+0^2}\\\Rightarrow v=30298514.82\ m/s$

The object would be moving at a velocity of 30298514.82 m/s

Answer 2

Answer:

Velocity will be $v=3.266\times 10^5m/sec$

Explanation:

We have given mass of the star $M=2\times 10^{30}kg[/tex}

Radius of the star [tex]R=5\times 10^{3}m$

Gravitational constant $G=6.67\times 10^{-11}Nm^2/kg^2$

We know that acceleration is given by $a=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times 2\times 10^{30}}{(5\times 10^3)^2}=5.33\times 10^{12}m/sec^2$

Displacement is given as s = 0.017 m

From third equation of motion

$v^2=u^2+2as$

As initial velocity u = 0 m/sec

So $v^2=0^2+2\times 5.33\times 10^{12}\times 0.017$

$v=3.266\times 10^5m/sec$