A planet of mass m moves in an elliptical orbit around an unknown star of mass M such that its maximum and minimum distances from the star are equal to r1 and r2 respectively. The angular momentum of the planet relative to the centre of the star is

Question

A planet of mass m moves in an elliptical orbit around an unknown star of mass M such that its maximum and minimum distances from the star are equal to r1 and r2 respectively. The angular momentum of the planet relative to the centre of the star is (A) m √(2GM r1r2/r1 + r2)  (B) 0 (C) m √(2 G M(r1+r2)/41 r2) (D) m √(2GM m r1/(r1 + r2) r2 )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c8ea8bbe17a1ce8d335862a2a61802bc-.png" /> According to the law of conservation of angular momentum, m V1 I1 =m V2 I2 ⇒ V2 =(V1 r1/r2) ......(i) From the law of conservation of total mechanical energy. (-G M m/r1)+(1/2) m v12=-(G M m/r2)+(1/2) m v22 ......(ii) From Eqs. (i) and (ii), we get V1=√(2 G M r2/(r1+r2) r1) Angular momentum, L=m v1 r1=m(√(2 G M r2/(r1+r2) r1)) × r1 L=m √(2 G M r1 I2/r1+r2)

Q. A planet of mass $m$ moves in an elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distances from the star are equal to $r_{1}$ and $r_{2}$ respectively. The angular momentum of the planet relative to the centre of the star is

$m \frac{2 GM r _{1} r _{2}}{r _{1} + r _{2}}$

0

$m \frac{2 GM ( r _{1} + r _{2} )}{4 _{1} r _{2}}$

$m \frac{2 GM m r _{1}}{( r _{1} + r _{2} ) r _{2}}$

Q. A planet of mass m moves in an elliptical orbit around an unknown star of mass M such that its maximum and minimum distances from the star are equal to r1​ and r2​ respectively. The angular momentum of the planet relative to the centre of the star is

mr1​+r2​2GMr1​r2​​​

0

m41​r2​2GM(r1​+r2​)​​

m(r1​+r2​)r2​2GMmr1​​​

Q. A planet of mass $m$ moves in an elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distances from the star are equal to $r_{1}$ and $r_{2}$ respectively. The angular momentum of the planet relative to the centre of the star is

$m \frac{2 GM r _{1} r _{2}}{r _{1} + r _{2}}$

$m \frac{2 GM ( r _{1} + r _{2} )}{4 _{1} r _{2}}$

$m \frac{2 GM m r _{1}}{( r _{1} + r _{2} ) r _{2}}$