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Question Number 29144 by tawa tawa last updated on 04/Feb/18

$$\mathrm{A}\mathrm{body}\mathrm{moves}\mathrm{in}\mathrm{a}\mathrm{circular}\mathrm{orbit}\mathrm{of}\mathrm{radius}4\mathrm{R}\mathrm{round}\mathrm{the}\mathrm{earth}.\mathrm{Express}\mathrm{the}\mathrm{acceleration}$$$$\mathrm{of}\mathrm{the}\mathrm{free}\mathrm{fall}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{g}$$$$\mathrm{R}=\mathrm{radius}\mathrm{if}\mathrm{the}\mathrm{earth}$$$$\mathrm{g}=\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}$$

Answered by ajfour last updated on 04/Feb/18

$$\frac{GMm}{{\left(4R\right)}^2}=ma$$$$\Rightarrow a=\frac{GM}{16R^2}=\frac{g}{16}.$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{Please}\mathrm{how}\mathrm{did}\mathrm{it}\mathrm{becomes}\frac{\mathrm{g}}{16}\mathrm{from}\frac{\mathrm{GM}}{{16\mathrm{R}}^2}$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{or}\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$$

Commented by ajfour last updated on 04/Feb/18

$$yes.$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

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