Question

Gravitational force between two object of mass 10kg and 15kg is 8×10^-8 N. Then find out separation between objects.

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Answer 1

Answer:

Take in both the masses and the distance between the masses and store it in separate variables.

2. Initialize one of the variables to the value of gravitational constant, G.

3. The the formula is used to determine the force acting between the masses.

4. Rounding off up-to two decimal places, print the value of the force.

5. Exit.

Answer 2

Step-by-step explanation:

Given :

Masses of the objects (m₁ and m₂) = 10 kg and 15 kg

Gravitational force of attraction (G) = 8 × 10⁻⁸ N

To Find :

The distance of seperation between the objects

Solution :

The Expression for gravitational force of attraction between two masses is given by,

$\begin{gathered} \\ \star \: {\boxed{\purple{\sf{F = \dfrac{Gm_1m_2}{ {r}^{2} }}}}} \\ \end{gathered} </p><p>⋆ </p><p>F= </p><p>r </p><p>2</p><p> </p><p>Gm </p><p>1</p><p> </p><p> m </p><p>2</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p>$

$\begin{gathered} \\ \\ \sf{where}\begin{cases}& \sf{G\:is\:Gravitational\:constant}\\ &\sf{m_1 \:is\:the\:mass\:of\:first\:object}\\ &\sf{m_2\:is\:the\:mass\:of\:another\:object} \\& \sf{r \: is \: distance \: of \: seperation \: between \: the \: masses} \end{cases}\\ \\\end{gathered} </p><p>where </p><p>⎩</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎨</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎧</p><p> </p><p> </p><p> </p><p> </p><p>GisGravitationalconstant</p><p>m </p><p>1</p><p> </p><p> isthemassoffirstobject</p><p>m </p><p>2</p><p> </p><p> isthemassofanotherobject</p><p>risdistanceofseperationbetweenthemasses</p><p> </p><p>$

Substituting the values in the formula ,

$\begin{gathered} \\ : \implies \sf \: 8 \times {10}^{ - 8} \: = \dfrac{(6.67 \times {10}^{ - 11}) (10 \: )(15 \: )}{ {r}^{2} } \\ \\ \end{gathered} </p><p>:⟹8×10 </p><p>−8</p><p> = </p><p>r </p><p>2</p><p> </p><p>(6.67×10 </p><p>−11</p><p> )(10)(15)$

$\begin{gathered} \\ : \implies \sf \: 8 \times {10}^{ - 8} = \frac{150\times 6.67 \times {10}^{ - 11} }{ {r}^{2} } \\ \\ \end{gathered} </p><p>:⟹8×10 </p><p>−8</p><p> = </p><p>r </p><p>2</p><p> </p><p>150×6.67×10 </p><p>−11$

$\begin{gathered} \\ : \implies \sf \: 8 \times {10}^{ - 8} \times {r}^{2} = 1000.5 \times {10}^{ - 11} \\ \\ \end{gathered} </p><p>:⟹8×10 </p><p>−8</p><p> ×r </p><p>2</p><p> =1000.5×10 </p><p>−11$

$\begin{gathered} \\ : \implies \sf \: {r}^{2} = \dfrac{1000.5 \times {10}^{ - 11} }{8 \times {10}^{ - 8} } \\ \\ \end{gathered} </p><p>:⟹r </p><p>2</p><p> = </p><p>8×10 </p><p>−8</p><p> </p><p>1000.5×10 </p><p>−11$

$\begin{gathered} \\ : \implies \sf \: {r}^{2} = 125 \times {10}^{ - 3} \\ \\ \end{gathered} </p><p>:⟹r </p><p>2</p><p> =125×10 </p><p>−3$

$\begin{gathered} \\ : \implies \sf \: r \: = \sqrt{125 \times {10}^{ - 3} } \\ \\ \end{gathered} </p><p>:⟹r= </p><p>125×10 </p><p>−3</p><p>$

$\begin{gathered} \\ : \implies{\underline{\boxed{\pink{\sf{ \: r = 0.35 \: m}}}}} \: \bigstar \\ \\ \end{gathered} </p><p>:⟹ </p><p>r=0.35m</p><p>$

★

Hence ,

The distance of seperation between them is 0.35 m

Note :

G = 6.67 × 10⁻¹¹ Nm²/kg²