AP EAPCET 2024 - 20th May Evening Shift | Gravitation Question 3 | Physics

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 )

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

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

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

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

AP EAPCET 2024 - 19th May Evening Shift

MCQ (Single Correct Answer)

-0

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ( $R=$ Radius of the earth)

$\frac{R}{2}$

$\frac{16 R}{9}$

$\frac{R}{3}$

$\frac{R}{8}$

AP EAPCET 2024 - 18th May Morning Shift

MCQ (Single Correct Answer)

-0

Two satellites of masses $m$ and 1.5 m are revolving around the earth with different speeds in two circular orbits of heights $R_E$ and $2 R_E$ respectively, where $R_F$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces on the earth due to the two satellites is

$2: 5$

$2: 3$

$1: 2$

$1: 5$

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