Question

$\huge\red{\mid{\fbox{\tt{Question:-}}\mid}}$

Two planets of mass M and 16M of radius a and 2a respectively, are at distance 10a. Find the minimum speed of a particle of mass m at surface of smaller planet so that it can reach from smaller planet to Larger planet.​

Answer 1

$\large\bold{\underline{\underline{Question:-}}}$

Two planets of mass M and 16M of radius a and 2a respectively, are at distance 10a. Find the minimum speed of a particle of mass m at surface of smaller planet so that it can reach from smaller planet to Larger planet.

$\large\bold{\underline{\underline{Answer:-}}}$

Option 1) v = √{5/9 × (GM/a)}

$\large\bold{\underline{\underline{Given:-}}}$

Two planets of mass M and 16M of radius a and 2a respectively.

They are at distance 10a.

$\large\bold{\underline{\underline{To \: Find:-}}}$

The minimum speed of a particle of mass at surface of smaller planet so that it can reach from smaller planet to Larger planet.

$\large\bold{\underline{\underline{Explanation:-}}}$

Let the distance from the point to the small planet where gravitational force of attraction of both the planets be equal be x.

$\leadsto\sf \dfrac{GMm}{x^2} = \dfrac{16GMm}{(10a-x)^2}$

$\leadsto\sf \dfrac{1}{x^2} = \dfrac{16}{(10a-x)^2}$

Take square root on both sides

$\leadsto\sf \dfrac{1}{x} = \dfrac{4}{10a-x}$

$\leadsto\sf 10a - x = 4x$

$\leadsto\sf 5x = 10a$

$\leadsto\sf x = 2a$

Total energy on the surface of the small planet:

Total energy = Potential energy + kinetic energy

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} - \dfrac{GMm}{a} -\dfrac{16GMm}{9a}$

a is the distance from the centre of the smaller planet to the surface of the smaller planet.

9a is the distance from the centre of the larger planet to the surface of the smaller planet.

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} + \dfrac{ - 9GMm - 16GMm}{9a}$

$\leadsto \sf E_i = \dfrac{1}{2}m {v}^{2} - \dfrac{ 25GMm}{9a}$

Total energy at the null point:

Total energy = Potential energy + kinetic energy

$\leadsto\sf E_f=0 - \dfrac{GMm}{2a} -\dfrac{16GMm}{8a}$

2a is the distance from the centre of the smaller planet to the null point.

8a is the distance from the centre of the larger planet to the null point.

$\leadsto\sf E_f= \dfrac{- 4GMm - 16GMm}{8a}$

$\leadsto\sf E_f= \dfrac{- 20GMm}{8a}$

$\leadsto\sf E_f= \dfrac{- 5GMm}{2a}$

$\boxed { \large \sf E_i= E_f}$

$\leadsto \sf \dfrac{1}{2}m {v}^{2} - \dfrac{ 25GMm}{9a} = \dfrac{- 5GMm}{2a}$

$\leadsto\sf \dfrac{1}{2}m {v}^{2} = \dfrac{- 5GMm}{2a} + \dfrac{ 25GMm}{9a}$

$\leadsto \sf \dfrac{1}{2} {v}^{2} = \dfrac{- 5GM}{2a} + \dfrac{ 25GM}{9a}$

$\leadsto \sf \dfrac{1}{2} {v}^{2} =\dfrac{- 45GM + 50GM}{18a}$

$\leadsto \sf {v}^{2} =\dfrac{5GM}{9a}$

$\leadsto \sf v = \sqrt{\dfrac{5}{9}\dfrac{GM}{a}}$

$\boxed{ \sf \green { \large{Hence \ the \ v_{min} = \sqrt{\dfrac{5}{9}\dfrac{GM}{a}}}}}$

Answer 2

$\huge\green {\mid{\fbox{\tt{Question:-}}\mid}}$

ᴛᴡᴏ ᴘʟᴀɴᴇᴛs ᴏғ ᴍᴀss ᴍ ᴀɴᴅ 16ᴍ ᴏғ ʀᴀᴅɪᴜs ᴀ ᴀɴᴅ 2ᴀ ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ, ᴀʀᴇ ᴀᴛ ᴅɪsᴛᴀɴᴄᴇ 10ᴀ. ғɪɴᴅ ᴛʜᴇ ᴍɪɴɪᴍᴜᴍ sᴘᴇᴇᴅ ᴏғ ᴀ ᴘᴀʀᴛɪᴄʟᴇ ᴏғ ᴍᴀss ᴍ ᴀᴛ sᴜʀғᴀᴄᴇ ᴏғ sᴍᴀʟʟᴇʀ ᴘʟᴀɴᴇᴛ sᴏ ᴛʜᴀᴛ ɪᴛ ᴄᴀɴ ʀᴇᴀᴄʜ ғʀᴏᴍ sᴍᴀʟʟᴇʀ ᴘʟᴀɴᴇᴛ ᴛᴏ ʟᴀʀɢᴇʀ ᴘʟᴀɴᴇᴛ.

Answer:-

• ᴏᴘᴛɪᴏɴ 1) ᴠ = {ʀᴏᴏᴛ 5/9 × (ɢᴍ / ᴀ)}

Given:-

• ᴛᴡᴏ ᴘʟᴀɴᴇᴛs ᴏғ ᴍᴀss ᴍ ᴀɴᴅ 16ᴍ ᴏғ ʀᴀᴅɪᴜs ᴀ ᴀɴᴅ 2ᴀ ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.
• ᴛʜᴇʏ ᴀʀᴇ ɪɴ ᴅɪsᴛᴀɴᴄᴇ 10 ᴀ.

To Find:-

• ᴛʜᴇ ᴍɪɴɪᴍᴜᴍ sᴘᴇᴇᴅ ᴏғ ᴀ ᴘᴀʀᴛɪᴄʟᴇ ᴏғ ᴍᴀss ᴀᴛ sᴜʀғᴀᴄᴇ ᴏғ sᴍᴀʟʟ ᴘʟᴀɴᴇᴛ sᴏ ᴛʜᴀᴛ ɪᴛ ɪs ᴄᴀɴ ʀᴇᴀᴄʜ ɪɴ sᴍᴀʟʟᴇʀ ᴘʟᴀɴᴇᴛ ᴛᴏ ʟᴀʀɢᴇʀ ᴘʟᴀɴᴇᴛ.

Explanation:-

ʟᴇᴛ ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴘᴏɪɴᴛ ᴛᴏ ᴛʜᴇ sᴍᴀʟʟ ᴘʟᴀɴᴇᴛ ᴡʜᴇʀᴇ ɢʀᴀᴠɪᴛᴀᴛɪᴏɴᴀʟ ғᴏʀᴄᴇ ᴏғ ᴀᴛᴛʀᴀᴄᴛɪᴏɴ ᴏғ ʙᴏᴛʜ ᴘʟᴀɴᴇᴛs ʙᴇ ᴇϙᴜᴀʟ ᴛᴏ ʙᴇ x.

$\frac{GMm}{ {x}^{2} } = \frac{16GMm}{(10x - x)} \\ \frac{1}{ {x}^{2} } \: = \frac{16}{(10x - {x}^{2} )}$