(a) Evaluate, in MeV, the energy of gravitational attraction for two spherically symmetrical protons with a center-to-center separation of 2 F.
(b) Do the same for the energy of Coulomb repulsion at that separation.
(c) Compare your results with the energy of nuclear attraction, which is about —10 MeV at that separation.

Explanation:

a ) Gravitational potential energy = - GMm/ r , G is universal gravitational constant , M and m are masses at distance r .

Putting the given values

Potential energy = -6.67 x 10⁻¹¹ x (1.67 x 10⁻²⁷)² / 2 x 10⁻¹⁵

=- 18.60 x 10⁻⁵⁰ J

=- 18.60 X 10⁻⁵⁰ / (1.6 x 10⁻¹⁹ x 10⁶) MeV

= - 11.62 x 10⁻³⁷ MeV

b ) electric potential energy = k q₁ x q₂ / r , k is constant which is equal to 9 x 10⁹ , q₁ and q₂ are charges  and r is distance between charges

putting the values for two proton

= 9 x 10⁹ x ( 1.6 x 10⁻¹⁹ )² / 2 x 10⁻¹⁵ J

= 11.52 x 10⁻¹⁴ J

= 11.52 x 10⁻¹⁴ / (1.6 X 10⁻¹⁹ X 10⁶) MeV

= + .72 MeV

c ) energy of nuclear attraction = - 10 MeV.

It is more than energy of nuclear repulsion by  more than 10 times.

answer and explanation;

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft