Question

A neutron star has about one and a half times the mass of our Sun but has collapsed to a radius of 10 kmkm . Part A What is the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface?

Answer 1

Answer:

gₓ = 1.36 x 10¹³ g

Explanation:

The value of acceleration due to gravity at a certain place is given by the following formula:

gₓ = GM/R²

where,

gₓ = acceleration due to gravity on the surface of neutron star

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the star = 10 * Mass of sun = (10)(2 x 10³⁰ kg) = 2 x 10³¹ kg

R = 10 km = 10⁴ m

Therefore,

gₓ = (6.67 x 10⁻¹¹ N.m²/kg²)(2 x 10³¹)/(10⁴)²

gₓ = 1.334 x 10¹⁴ m/s²

Hence, comparing it with the free-fall acceleration at Earth's Surface:

gₓ/g = (1.334 x 10¹⁴)/9.8

gₓ = 1.36 x 10¹³ g

Answer 2

The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is $1.35 \times 10^5 \ g_E$.

The given parameters:

• Mass of the neutron star, m = 1.5 M
• Radius of the neutron star, R = 10 kmkm

The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is calculated as follows;

$F = mg = \frac{GM_sm}{R^2} \\\\(1.5 M_s)g = \frac{GM_s(1.5 M_s)}{R^2} \\\\g = \frac{GM_s}{R^2}$

where;

$M_s$ is the mass of the Sun = 1.989 x 10³⁰ kg.

$g = \frac{6.67 \times 10^{-11} \times 1.989 \times 10^{30} }{(10,000,000)^2} \\\\g = 1.326 \times 10^{6} \ m/s^2$

In terms of gravity of Earth $(g_E)$;

$= \frac{1.326 \times 10^6}{9.81} = 1.35 \times 10^5 \\\\= 1.35 \times 10^5 \ g_E$

Thus, the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is $1.35 \times 10^5 \ g_E$.

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