Assume that the mass of a neutron star is 1.5 times that of the Sun and the radius of the neutron star is 10 km. Ignore effects of Special Relativity.

What is the approximate speed at which an object released from far away (at zero speed) reaches the surface of a neutron star?
A. 8*10^7 m/s
B. 9*10^7 m/s
C. 2*10^8 m/s
D. 3*10^8 m/s
Thank you.

Hi, Elena. Thanks for helping me with the neutron star problem. I knew the formula, but kept getting answers that were not close to the possible choices. What variables did you use. Thanks again for your help.

v=sqrt{2•6.67•10⁻¹¹•1.5•1.98•10³º/10⁴}=

=1.99•10⁸ m/s ≈2•10⁸ m/s

To calculate the approximate speed at which an object released from far away reaches the surface of a neutron star, we can use the law of conservation of mechanical energy. The mechanical energy of an object falling towards a massive body is given by the equation:

E = K + U

where E is the total mechanical energy, K is the kinetic energy, and U is the gravitational potential energy.

The gravitational potential energy of an object near the surface of a massive body is given by:

U = -GMm / r

where G is the gravitational constant, M is the mass of the massive body, m is the mass of the object, and r is the distance between the object and the center of the massive body.

The kinetic energy of the object can be calculated using the equation:

K = (1/2)mv^2

where v is the velocity of the object.

Since the object is released from far away with zero velocity, its initial kinetic energy is zero.

At the surface of the neutron star, the radius (r) is given as 10 km.

Given that the neutron star's mass (M) is 1.5 times that of the sun's mass, we can calculate it as:

M = 1.5 * mass of the sun

To solve for the velocity (v) of the object at the surface of the neutron star, we need to equate the initial mechanical energy (E) to the final mechanical energy (E') at the surface of the neutron star:

E = K + U = E'

Since the initial kinetic energy (K) is zero, the equation becomes:

0 + U = E'

Setting the two equations for gravitational potential energy equal to each other, we have:

(-GMm / r) = (-(GMm' / R))

where m' is the mass of the neutron star, G is the gravitational constant, and R is the radius of the neutron star.

Since we are ignoring relativistic effects, we can solve for the velocity (v) by solving for m' and substituting it into the equation:

m' ≈ M ≈ 1.5 * mass of the sun

Substituting the values into the equation, we get:

(-GMm / r) = (-(GM(1.5 * mass of the sun) / R))

Now we can solve for the velocity (v) using the equation for gravitational potential energy:

v = sqrt(2G(1.5 * mass of the sun) / R)

Substituting the given values of G, mass of the sun, and the radius of the neutron star, we can calculate the approximate speed at which the object reaches the surface:

v ≈ sqrt(2 * 6.67430 * (10^-11) * (1.5 * 1.989 * (10^30)) / 10000)

Simplifying the equation, we get:

v ≈ 2 * 10^8 m/s

Therefore, the approximate speed at which the object reaches the surface of the neutron star is approximately 2 * 10^8 m/s, which corresponds to option C.

Please note that this is an approximate value and that the calculation does not take into account the effects of Special Relativity.

C.

M=1.5•1.98•10³º kg
G =6.67•10⁻¹¹ N•m²/kg²
mv²/2=GmM/R
v=sqrt{2GM/R}=2•10⁸ m/s