AP EAPCET 2024 - 21th May Morning Shift | Gravitation Question 3 | Physics

The time period of revolution of a satellite $T$ around the carth depends on the radius of the circular orbit $R$. mass of the earth $M$ and universal gravitational constant $G$. The expression for $T$, using dimensional analysis is ( $K$ is constant of proportionality)

$K \sqrt{\frac{R^2}{G M}}$

$K \sqrt{\frac{R}{G M}}$

$K \sqrt{\frac{R^3}{G M}}$

$K \sqrt{\frac{R^3}{G M^2}}$

AP EAPCET 2024 - 21th May Morning Shift

MCQ (Single Correct Answer)

-0

If the time period of revolution of a satellite is $T$, the its kinetic energy is proportional to

$T^{-1}$

$T^{-2}$

$T^{-3}$

$T^{-2 / 3}$

AP EAPCET 2024 - 20th May Evening Shift

MCQ (Single Correct Answer)

-0

What is the height from the surface of earth, where acceleration due to gravity will be $1 / 4$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$

6400 km

3200 km

1600 km

640 km

AP EAPCET 2024 - 20th May Morning Shift

MCQ (Single Correct Answer)

-0

The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5 \mathrm{~ms}^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth= 6400 km )

$1.11 \mathrm{~ms}^{-2}$

$1.5 \mathrm{~ms}^{-2}$

$2.22 \mathrm{~ms}^{-2}$

$1.25 \mathrm{~ms}^{-2}$

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