MHT CET 2023 12th May Morning Shift | Gravitation Question 14 | Physics

Consider a light planet revolving around a massive star in a circular orbit of radius '$$r$$' with time period '$$T$$'. If the gravitational force of attraction between the planet and the star is proportional to $$\mathrm{r}^{\frac{7}{2}}$$, then $$\mathrm{T}^2$$ is proportional to

$$r^{9 / 2}$$

$$r^{7 / 2}$$

$$r^{5 / 2}$$

$$r^{3 / 2}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The radius of the orbit of a geostationary satellite is (mean radius of earth is '$$R$$', angular velocity about own axis is '$$\omega$$' and acceleration due to gravity on earth's surface is '$$g$$')

$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$$

$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}$$

$$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}$$

$$\frac{\mathrm{gR}^2}{\omega^2}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The ratio of energy required to raise a satellite to a height '$$h$$' above the earth's surface to that required to put it into the orbit at the same height is ($$\mathrm{R}=$$ radius of earth)

$$\frac{2 \mathrm{~h}}{\mathrm{R}}$$

$$\frac{h}{R}$$

$$\frac{\mathrm{R}}{\mathrm{h}}$$

$$\frac{\mathrm{R}}{2 \mathrm{~h}}$$

MHT CET 2023 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The radius of earth is $$6400 \mathrm{~km}$$ and acceleration due to gravity $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$. For the weight of body of mass $$5 \mathrm{~kg}$$ to be zero on equator, rotational velocity of the earth must be (in $$\mathrm{rad} / \mathrm{s}$$ )

$$\frac{1}{80}$$

$$\frac{1}{400}$$

$$\frac{1}{800}$$

$$\frac{1}{1600}$$

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