Question

Calculate the acceleration due to gravity on the surface of satellite having
mass 7.4 x 10^22kg and radius 1.74 x 10^6cm. (G = 6.7 x 10^–11Nm/kg^2)

Answer 1

Answer:

I am confused so confusing

Answer 2

Explanation:

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration=16.3\times10^{2}\:m/s^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given:}}} \\ \tt: \implies Mass \: of \: satellite(M) = 7.4 \times {10}^{22} \: kg \\ \\ \tt: \implies Radius(r) = 1.74 \times {10}^{6} \: cm \\ \\ \tt: \implies G = 6.67 \times {10}^{ - 11} \: Nm/ {kg}^{2} \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies Acceleration \: due \: to \: gravity(g) =?$

• According to given question :

$\tt \circ \: R= \frac{1740000}{100} =1.74 \times {10}^{4} \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies g = \frac{GM}{ {R}^{2} } \\ \\ \tt: \implies g = \frac{6.67 \times {10}^{ - 11} \times 7.4 \times {10}^{22} }{ {(1.74 \times {10}^{4} )}^{2} } \times \frac{N \times m \times kg}{ {kg}^{2} \times {m} } \\ \\ \tt: \implies g = \frac{6.67 \times 7.4 \times {10}^{ - 11 + 22} }{ {(1.74)}^{2} \times {10}^{8} } \times \frac{m}{ {s}^{2} } \\ \\ \tt: \implies g = \frac{49.358 \times {10}^{10} }{3.0276 \times {10}^{8} } \times m/{s}^{2} \\ \\ \green{\tt: \implies g = 16.3 \times {10}^{2} \: m/{s}^{2} }\\\\\green{\tt\therefore Acceleration\:due\:to\:gravity\:is\:16.3\times10^{2}\:m/s^{2}}$