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?

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

i think the answer would be reserve.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y