Question

Assuming that earth is a sphere of uniform density 5.5gm/cc and radius 6.4×10 8 cm.Find the value of 'g' on the surface of the earth​

Answer 1

Correct Question:

Assuming the earth to be a uniform sphere of radius 6400km and density 5.5 g/c.c, find the value of g on its surface.$\bold{G=6.66×10^{-11} Nm^2kg^{-2}}$.

Answer:

• The value of 'g' on the surface of the earth is 9.8148864 m/s².

Explanation:

Given that,

$\bull \sf \: Radius (R) = 6400 km \leadsto \: 6.4 \times 10 {}^{6} m \\ \\ </h3><h3> \bull \sf \: Density \bold{(\rho)} = 5.5 g/c.c \leadsto \: 5.5 \times 10 {}^{3}kg/m {}^{3} \\ \\ </h3><h3> \bull\sf{G=6.66×10^{-11} Nm^2kg^{-2}}$

As we know that,

$\red \bigstar \boxed{\sf \: g = \frac{4}{3}πGR \rho</h2><h2>}$

[ Putting values ]

$\leadsto \sf \: g = \frac{4}{3} \times 3.14 \times 6.66 \times 10 {}^{ - 11} \times 6.4 \times 10 {}^{6} \times 5.5 \times 10 {}^{3} \\ \\ \leadsto \sf \: g = 4 \times 3.14 \times 2.22 \times 6.4 \times 5.5 \times 10 {}^{ - 11 + 6 + 3} \\ \\ \leadsto \sf \: g = 27.8832 \times 35.2 \times 10 {}^{ - 2} \\ \\ \leadsto \sf g = 981.48864 \times 10 { }^{ - 2} \\ \\ \leadsto \sf g = 9.8148864 \: ms {}^{ - 2} \: \: \green \bigstar$

Answer 2

Answer:

sorry am not answer very very sorry