Question

An object projected with a velocity greater than or equal to 11.2 km s⁻¹ will not return to earth. Explain the reason.

Answer 1

Escape speed is that after attaining which a body moves just out of the earth’s Gravitational influence. and it is possible only when total energy of body is zero or positive.

now, we know the Total energy of a body is the sum of kinetic energy and potential energy, T = K + U, Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body which is always positive and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet.

Kinetic Energy is given as K = 1/2 mv²

The potential energy of a body above the surface of the earth is given as U = -GMm/R


If Body have to move out of Earth’s influence its total energy should be positive i.e. T ≥ 0

or, 1/2 mv² - GMm/R ≥ 0

or, 1/2 mv² ≥ GMm/R

or, v² ≥ 2GM/R

or, v ≥ √{2GM/R}

hence, v = √{2GM/R} is the minimum speed given to a body , by which cannot come back into the earth. because they just out from gravitational influence by this speed.

after putting, G = 6.67 × 10^-11 Nm²/kg², M = 6 × 10²⁴ Kg and R = 6.4 × 10^6m
we get, v ≈ 11.2 km/s

hence, An object projected with a velocity greater than or equal to 11.2 km/s will not return to earth

Answer 2

Explanation:

Let initial force of attraction between two objects is F when distance between objects is r.

But force of attraction decreased by 36% when the distance between them is increased by 4.

e.g., final force = F - 36% F = 0.64F

We know, gravitational force is inversely proportional to square of distance between objects.

so, \frac{F_1}{F_2}=\frac{d_2^2}{d_1^2}

F

2

F

1

=

d

1

2

d

2

2

or, \frac{F}{0.64F}=\frac{(r+4)^2}{r^2}

0.64F

F

=

r

2

(r+4)

2

or, 1/(0.8)² = (r + 4)/r

or, 1/ 0.8 = (r + 4)/r

or, r = 0.8r + 3.2

or, r = 16 m

hence, original distance between them is 16m