Problem: Find the minimum distance of a planet from the sun given that the maximum and minimum velocities are 3×10^4 and 4×10^3 respectively and the maximum distance of the planet from the sun is 4×10^4 km.

Solution:

We can use the conservation of energy principle to solve the problem. The potential energy of the planet is given by the formula:

U(r) = -GMm/r,

where G is the gravitational constant, M and m are the masses of the sun and the planet respectively, and r is the distance between them.

The kinetic energy of the planet is given by the formula:

K(v) = 1/2mv^2,

where m is the mass of the planet and v is its velocity.

The total energy of the planet is given by the sum of its kinetic and potential energies:

E = K(v) + U(r)

Since energy is conserved, the total energy of the planet is constant. Therefore, we can equate the total energy at the minimum distance to the total energy at the maximum distance:

K(max) + U(max) = K(min) + U(min)

1/2m(3×10^4)^2 - GMm/4×10^4 = 1/2m(4×10^3)^2 - GMm/r

Solving for r, we get:

r = GM/[(3×10^4)^2/2 - (4×10^3)^2/2 + 4×10^4]

r = GM/[(9×10^8) - (8×10^6) + 4×10^4]

r = GM/(8.92×10^8)

We know that the maximum distance of the planet from the sun is 4×10^4 km. Therefore, we can use the equation of an ellipse to find the minimum distance:

a = (r + 4×10^4)/2

b = sqrt[(r)(4×10^4)]/2

c = (a^2 - b^2)^(1/2)

where a is the semi-major axis, b is the semi-minor axis, and c is the distance between the center and one of the foci.

The minimum distance of the planet from the sun is given by:

d = a - c

Substituting the values of a, b, and c, we get:

d = [(r + 4×10^4)/2] - [(r^2)(4×10^4)]^(1/2)/2

Substituting the value of r, we get:

d = [(GM/8.92×10^8 + 4×10^4)/2] - [(GM^2/8.92×10^8)^2(4×10^4)]^(1/2)/2

Therefore, the minimum distance of the planet from the sun is a function of the gravitational constant G and the masses of the sun and the planet.

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