The distance between sun and a planet is r. The
angular momentum of planet around the sun in
circular orbit is proportional to
Answers
Answer:
Answer:
Angular momentum, L\propto r^{1/2}.L∝r
1/2
.
Explanation:
Given that, rr is the radial distance between the sun and the planet.
The angular momentum of the planet around the sun is given by
L=mvrL=mvr .
where,
mm = mass of the planet.
vv = speed of the planet.
The speed of the planet also varies with rr .
The centripetal force with which the planet is revolving around the sun must be balanced with the gravitational force on the planet due to sun.
The centripetal force is given by
F_c = \dfrac{mv^2}{r}F
c
=
r
mv
2
.
The Gravitational force on the planet due to Sun is given by
F_g =\dfrac{GMm}{r^2}F
g
=
r
2
GMm
.
where,
GG = Universal Gravitational constant.
MM = mass of the Sun.
On putting these two equal,
\begin{lgathered}F_e=F_g\\\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}\\v^2=\dfrac{GM}{r}\\v=\sqrt{\dfrac{GM}{r}}\end{lgathered}
F
e
=F
g
r
mv
2
=
r
2
GMm
v
2
=
r
GM
v=
r
GM
Putting this value of speed in the expression of angular momentum, we get,
\begin{lgathered}L=mvr=m\left ( \sqrt{\dfrac{GM}{r}} \right )r=\left(m\sqrt{GM \right) r^{1/2}\\\Rightarrow L\propto r^{1/2}.\end{lgathered}
IF you satisfied with my answer so mark me as brainliest.
Answer:
the momentum is directly proportional to its weight