A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius (R/2), and the other mass, in a circular orbit of radius (3R/2). The difference between the final and initial total energies is :

Question

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius (R/2), and the other mass, in a circular orbit of radius (3R/2). The difference between the final and initial total energies is : (A)  - (GM m/2R) (B)  + (GM m/6R) (C)  (GM m/2R) (D)  - (GM m/6R)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7eafc6cc3c8aa296e1a417837798a854-.png" /> Initial energy =(-G M m/R)+(1/2) (G M m/R)=(-G M m/2 R) Final Energy =(-G M(m / 2)/R / 2)-(G M(m / 2)/3 R / 2)+(1/2) (G M(m / 2)/R / 2)+(1/2) (G M(m / 2)/3 R / 2) =(-G M(m / 2)/2 R / 2)-(G m(m / 2)/2 × 3 R / 2) Therefore, Δ E= Final energy - Initial energy =(-G M(m / 2)/2 R / 2)-(G M(m / 2)/2 × 3 R / 2)-((-G M m/2 R)) ⇒ Δ E=(-G M m/6 R)

$- \frac{GM m}{2 R}$

$+ \frac{GM m}{6 R}$

$\frac{GM m}{2 R}$

$- \frac{GM m}{6 R}$

−2RGMm​

+6RGMm​

2RGMm​

−6RGMm​

Initial energy =R−GMm​+21​RGMm​=2R−GMm​ Final Energy =R/2−GM(m/2)​−3R/2GM(m/2)​+21​R/2GM(m/2)​+21​3R/2GM(m/2)​ =2R/2−GM(m/2)​−2×3R/2Gm(m/2)​ Therefore, ΔE= Final energy − Initial energy =2R/2−GM(m/2)​−2×3R/2GM(m/2)​−(2R−GMm​) ⇒ΔE=6R−GMm​

$- \frac{GM m}{2 R}$

$+ \frac{GM m}{6 R}$

$\frac{GM m}{2 R}$

$- \frac{GM m}{6 R}$