Question

What is the magnitude of the gravitational force between the earth and a 1kg object on its surface ? (Mass of the earth is 6 × 10²⁴kg and radius of the earth is 6.4 × 10⁶m , (G = 6.7 × 10‐¹¹ Nm²kg-²) ?​

Answer 1

$\fcolorbox{cyan}{orange}{Solutions:-}$

$\fcolorbox{cyan}{pink}{Given:-}$

$Mass \: of \: the \: object \: on \: the \: Earth's \: surface \: (m _{1}) = 1 \: kg \: \\ Mass \: of \: Earth \: (m _{2}) = 6 \times {10}^{24} kg \\ Radius \: of \: Earth(R) = 6.4 \times {10}^{6} - - - - take \: it \: 6 \times {10}^{6} \: ( for \: more \: convenient)\\ Value \: of \: Universal \: gravitational \: constant(G) = 6.67 \times {10}^{ - 11} {kg}^{ - 2} \: N \: {m}^{2}$

$\fcolorbox{cyan}{green}{To find:-}$

• Magnitude of Gravitational force between the earth and 1kg object on its surface(F)

$\fcolorbox{cyan}{pink}{Formula:-}$

$F =G \frac{ m_{1} m_{2} }{ {R}^{2} }$

(Where F is the Magnitude of gravitational force between two bodies of masses $m_{1}$and $m_{2}$, G is the gravitational constant and R is the distance between two bodies)

$\fcolorbox{cyan}{purple}{By \: the \: problem:-}$

$Plot \: the \: value \: in \: the \: formula \: \\ F =G \frac{ m_{1} m_{2} }{ {R}^{2} } \\ F = \: \frac{(6.67 \times {10}^{ - 11}) \times (1) \times (6 \times {10}^{24} }{ {(6 \times {10}^{6})}^{2} } \\ F = \frac{6.67 \times {10}^{ - 11} \times 6 \times {10}^{24}}{6 \times {10}^{6} \times 6 \times {10}^{6} } \\ F = \frac{6.67 \times {10}^{13} }{6 \times {10}^{12} } \\ F = 1.11 \times 10 \\ F = 11 \: n$

$\fcolorbox{cyan}{green}{Answer:-}$

Therefore ,there will be a gravitational force of magnitude 11 Newton between the earth and 1 kg object.

Answer 2

${\huge{\underline{\underline{\bf{\maltese\; Question:-}}}}}$

Q - What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface ?

(Mass of the Earth - 6 × 10²⁴ kg and Radius of Earth - 6.4 × 10⁶ m )

( G = 6.7 × 10-¹¹ Nm²kg-²)

${\huge{\underline{\underline{\bf{\maltese\;Answer\checkmark}}}}}$

★ Concept :

• Gravitational force is a force that one body applies on the other to attract .

• This force is directly proportional to the product of masses of the two bodies .

• This is inversely proportional to the square of distance between the two objects .

★ We Have :

• Mass of Earth (m1) = 6 × 10²⁴ kg
• Mass of the other object (m2) = 1 kg
• Distance between m1 and m2 = radius of earth = 6.4 × 10⁶ m
• G = $6.7×{10}^{-11}$

★ To Find :

• Magnitude of Gravitational force (F)

★ Solution :

Using the Formula ↓

$\boxed{\rm\;F=G\frac{m_{1}m_{2}}{r^2}}$

Now , just put the values without any change because all the values here , are given in the S.I unit .

$\mapsto \rm \: f = 6.7 \times {10}^{ - 11} \: \frac{n \cancel{m}^{2} }{ \cancel{kg}^{2} } \frac{6 \times {10}^{24} \: \cancel{kg }\times 1 \: \cancel{kg} }{({6.4 \times {10}^{6})}^{2} \cancel{m}^{2} }$

$\mapsto \rm \: f = \frac{6.7 \times 6 \times {10}^{24 - 11} }{40.96 \times {10}^{12} } newtons$

$\mapsto \rm \: f = \frac{40.2 \times {10}^{13 - 12} }{40.96} \: newtons$

$\mapsto \rm \: f = 0.981 \times 10 \: newtons$

$\large \mapsto \boxed{ \rm \: f = 9.81 \: newtons}$

So ,

The magnitude of Gravitational force here would be 9.81 Newtons .