Question

What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface (radius of the Earth is 6.4*10-m)

Answer 1

Correct Question:-

What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface (radius of the Earth is 6400 km).

Given data:-

→ Mass of object (m) = 1 kg

→ Radius of Earth = 6400 km

Solution:-

We use formula to calculate gravitational force between the Earth and a 1 kg object on its surface.{we know that object on surface of earth hence distance between object and earth is equal to earths radius}

Gravitational foce between two particles denoted by F

${ \huge{ \purple {\dashrightarrow{\sf{F = G\frac{Mm}{ {R}^{2} } }}}}}$

Here,

→ G = gravitational constant

→ R = radius of earth

→ M = mass of earth

→ m = mass of object

→ F = Gravitational foce between two particles.

We know that,

→ G = 6.673 × 10^-11 Nm²/kg²

→ M = 5.972 × 10^24 kg.

Now,

${ \purple {\dashrightarrow{\sf{F = 6.673 \times {10}^{ - 11 } \times \frac{5.972 \times {10}^{24} \times 1}{ {(6400)}^{2} } }}}}$

${ \purple {\dashrightarrow{\sf{F = \frac{3.9851 \times {10}^{14} }{ 4096 \times {10}^{4} } }}}}$

${ \purple {\dashrightarrow{\sf{F = 9729248.047 \: \: N}}}}$

Hence, force between the Earth and a 1 kg object is 97.29748.047 N.

Answer 2

$\begin{gathered}⠀⠀⠀⠀\begin{gathered} \\ { { {\boxed {\rm {\blue { F = \dfrac{Gm_1m_2}{{r}^{2} }}}}}}} \\ \\ \\ \bf \: According \: to \: formula, \: \\ \\ \tt { \: F = \dfrac{6.7 \times {10}^{ - 11 } \times 6 \times {10}^{24} \times 1}{( \: 6.4 \times {10}^{6} ) {}^{2} } } \\ \\ \tt { \: F = \dfrac{ 6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} } } \\ \\ \tt { \: F = \dfrac{6.7 \times6 \times {10}^{( - 11 + 24)} }{6.4 \times 6.4 \times {10}^{(6 + 6)} } } \\ \\ \tt { \: F = \dfrac{6.7 \times 6 \times {10}^{13} }{6.4 \times 6.4 \times {10}^{12} } } \\ \\ \tt { \: F = \dfrac{40.2 }{40.96} \times {10}^{(13 - 12)} } \\ \\ \tt { \: F = \dfrac{40.2 }{40.96} \times 10 } \\ \\ \tt { \: F = \dfrac{402 \: \times \: \cancel{ 10 } \: \times 100}{4096 \: \times \: \cancel{10}} } \\ \\ \tt { \: F = \cancel {\dfrac{40200}{4096} } = 9.8 \: N} \\ \\ \\ \end{gathered} \end{gathered}$

Therefore , magnitude of the gravitational force between the earth and 1 kg object on its surface is 9.8 Newtons