MHT CET 2020 16th October Evening Shift | Gravitation Question 101 | Physics

The ratio of energy required to raise a satellite of mass $$m$$ to a height $$h$$ above the earth's surface of that required to put it into the orbit at same height is [$$R=$$ radius of the earth]

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

$$\frac{3 h}{R}$$

$$\frac{4 h}{R}$$

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

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

As we go from the equator of the earth to pole of the earth, the value of acceleration due to gravity

decreases

decreases up to latitude of $$45^{\circ}$$ and increases thereafter

remains same

increases

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

The mass of earth is 81 times the mass of the moon and the distance between their centres is $$R$$. The distance from the centre of the earth, where gravitational force will be zero is

$$\frac{R}{4}$$

$$\frac{R}{2}$$

$$\frac{9 R}{10}$$

$$\frac{R}{81}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

A body is thrown from the surface of the earth velocity $$\mathrm{v} / \mathrm{s}$$. The maximum height above the earth's surface upto which it will reach is ($$R=$$ radius of earth, $$g=$$ acceleration due to gravity)

$$\frac{v R}{2 g R-v}$$

$$\frac{2 g A}{v^2(R-1)}$$

$$\frac{v R^2}{g R-v}$$

$$\frac{v^2 R}{2 g R-v^2}$$

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