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

30298514.82 m/s

Step-by-step 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(M)/(r^2)\\\Rightarrow a=6.67* 10^(-11)(2* 10^(30))/(5* 10^3)\\\Rightarrow a=2.7* 10^(16)\ m/s^2`

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=\sqrt{2* 2.7* 10^(16)* 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

Velocity will be
`v=3.266* 10^5m/sec`

Step-by-step explanation:

We have given mass of the star
`M=2* 10^(30)kg[/tex}</p><p>Radius of the star [tex]R=5* 10^(3)m`

Gravitational constant
`G=6.67* 10^(-11)Nm^2/kg^2`

We know that acceleration is given by
`a=(GM)/(R^2)=(6.67* 10^(-11)* 2* 10^(30))/((5* 10^3)^2)=5.33* 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* 5.33* 10^(12)* 0.017`

`v=3.266* 10^5m/sec`