Question

derive a relation between accelaeation due to gravety and universal gravitational constant​

Answer 1

✯ ʀᴇʟᴀᴛɪᴏɴ ʙᴇᴛᴡᴇᴇɴ ɢʀᴀᴠɪᴛᴀᴛɪᴏɴᴀʟ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴀɴᴅ ᴜɴɪᴠᴇʀsᴀʟ ɢʀᴀᴠɪᴛᴀᴛɪᴏɴᴀʟ ᴄᴏɴsᴛᴀɴᴛ ✯

➪let an object of mass 'm' is placed on the earth

here ,

• mass of the earth = M (kg)
• radius of earth = r (km)
• Gravitational acceleration = g (m/s²)

★ by Newton's second law

☞︎︎︎force is equal to product of mass and acceleration

$f = mg ............(i)$

★ by Newton's law of gravitation

$f = \frac{Gm1m2}{ {r}^{2} }$

here,

• G = gravitational constant
• m1 = M (mass of the earth)
• m2 = m (mass of the object)
• r = r (radius of the earth)

then,

$f = \frac{GMm}{ {r}^{2} } ..............(ii)$

now, from equation (i) and (ii)

$\: \: \: \: \: \: \: \: \: \: f = f \\ \\ : \implies \:mg = \frac{GMm}{ {r}^{2} } \\ \\ : \implies \: \cancel{m}g = \frac{GM \cancel{m}}{ {r}^{2} } \\ \\ : \implies \: g = \frac{GM}{ {r}^{2}} \: \: \: \: \: \: \:$

☆ hence, relation between gravitational acceleration (g) and universal gravitational constant is,

$\bold{g = \frac{GM}{ {r}^{2}}}$