Question 448: Gravitation

Question

A geostationary satellite is orbiting the Earth at a height of $$4 R$$ above that surface of the Earth. $R$ being the radius of the earth. The, time period of another stellite in coins at a height of $$2 R$$ from the surface of the Earth is.

Accepted Answer

According to Kepler's law of period's $$T^2 \propto R^3$$
$$\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}=\frac{(4 R)^3}{(2 R)^3}=\frac{64 R^3}{8 R^3}=8 	ext { or } T_2=T_1 / 2 \sqrt{2}$$

Answer

$$2 T_1$$

Answer

$$2 \sqrt{2} T_1$$

Answer

$$T_1 / 2$$

Answer

$$T_1 / 2 \sqrt{2}$$

Gravitation

A geostationary satellite is orbiting the Earth at a height of $4 R above that surface of the Earth. R being the radius of the earth. The, time period of another stellite in coins at a height of 2 R$ from the surface of the Earth is.

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