Respuesta :
Answer:
[tex]331665750000\ m/s^2[/tex]
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
[tex]g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2[/tex]
The gravitational acceleration at the surface of such a star is [tex]331665750000\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s[/tex]
The velocity of the object would be 3257806.62409 m/s
(a) the acceleration due to gravity is 3.3 × 10¹¹ m/s²
(b) the final velocity of the object is 3.26 × 10⁶ m/s
Gravitational force:
Given that the neutron star has a mass equal to Sun, M = 2 × 10³⁰ kg
Radius of the star is R = 20 km = 20 × 10³
The weight on the surface of a neutron star that has the same mass as our Sun and a diameter of 21.0 km will be 8.29 × 10¹³ N
(a) The gravitational force on the surface of the neutron star is given by:
F = GMm/R²
mg = GMm/R²
where G is the gravitational constant
M is the mass of the body
m is the mass of the person
and, R is the radius of the body
So,
g = GM/R²
g = (6.67 × 10⁻¹¹)( 2 × 10³⁰)/ (210 × 10³)²
g = 3.3 × 10¹¹ m/s²
(b) According to the third equation of motion:
v² = u² + 2gh
v² = 2gh , since u = 0 as the object is at rest initially
v = [tex]\sqrt{2\times3.3\times10^{11}\times16}[/tex]
v = 3.26 × 10⁶ m/s
Learn more about gravitational force:
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