Question

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Answer 1

Answer:

$F = 7.723066756 \times {10}^{ - 9} \: N$

Explanation:

Given:

$There \: are \: 2 \: Gold \: Spheres \: say \: G1 \: and \: G2 \: having \: radius \: r1 = 20 \: cm = 0.2 \: m \: and \: r2 = 40 \: cm = 0.4 \: m \: respectively. \\ </em></p><p><em>[tex]There \: are \: 2 \: Gold \: Spheres \: say \: G1 \: and \: G2 \: having \: radius \: r1 = 20 \: cm = 0.2 \: m \: and \: r2 = 40 \: cm = 0.4 \: m \: respectively. \\ Density \: of \: Gold = 19.3 \frac{g}{ {cm}^{3} } = 19.3 \frac{ \frac{1kg}{1000} }{ { \frac{1m}{100} }^{3} } = 19.3 \times {10}^{3} \frac{kg}{ {m}^{3} } </em></p><p><em>[tex]There \: are \: 2 \: Gold \: Spheres \: say \: G1 \: and \: G2 \: having \: radius \: r1 = 20 \: cm = 0.2 \: m \: and \: r2 = 40 \: cm = 0.4 \: m \: respectively. \\ Density \: of \: Gold = 19.3 \frac{g}{ {cm}^{3} } = 19.3 \frac{ \frac{1kg}{1000} }{ { \frac{1m}{100} }^{3} } = 19.3 \times {10}^{3} \frac{kg}{ {m}^{3} } \\ Also, \: both \: spheres \: are \: separated \: by \: 170 \: m. \\ R = 170 \: m$

Solution:

$Volume \: of \: Sphere = \frac{4}{3} \pi {r}^{3}$

Now,

$Volume \: of \: Sphere \: G1 = \frac{4}{3} \pi {20}^{3} = \frac{16 \times \times 32000\pi}{3} \: {cm}^{3} \\ Volume \: of \: Sphere \: G2 = \frac{4}{3} \pi {40}^{3} = \frac{256000\pi}{3} \: {cm}^{3}$

We know,

$density = \frac{mass}{volume}$

Therefore,

$Mass = density \times volume$

$Let \: M1 \: and \: M2 \: be \: mass \: of \: gold \: sphere \: G1 \: and \: G2 \: respectively.$

Therefore,

$M1 = (density \: of \: gold) \times (Volume \: of \: Sphere \: G1) \\ M1 = (19.3 \frac{gm}{ {cm}^{3} } ) \times ( \frac{32000\pi}{3} {cm}^{3} ) = (19.3)( \frac{32000\pi}{3} ) \: gm = (19.3)( \frac{32\pi}{3} )kg. \\ M2 = (density \: of \: gold) \times (Volume \: of \: Sphere \: G2) \\ M2 = (19.3 \frac{gm}{ {cm}^{3} } ) \times ( \frac{256000\pi}{3} {cm}^{3} ) = (19.3)( \frac{256000\pi}{3} ) \: gm = (19.3)( \frac{256\pi}{3} )kg.$

$Now, \\ Gravitational \: force \: of \: attraction \: between \: two \: objects \: is \: given \: by: \\ F = \frac{G(M1)(M2)}{ {R}^{2} } \\ where \: G \: is \: gravitational \: constant \: and \: has \: value: \\ G = 6.67 \times {10}^{ - 11} \\ F = \frac{(6.67 \times {10}^{ - 11})((19.3)( \frac{32\pi}{3} )kg)((19.3)( \frac{256\pi}{3} )kg)}{ {170 \: m}^{2} } \\ F = 772.3066756 \times {10}^{ - 11} \: N \\ F = 7.723066756 \times {10}^{ - 9} \: N$