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Gravitation MCQs for NEET — Physics Questions with Answers

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If the radius of earth's orbit is made 1/4 th, the duration of an year will become -

8 times

4 times

1/8 times

1/4 times

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Explanation

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If mass of a satellite is doubled and time period remain constant, the ratio of orbit in the two cases will be

1 : 2

1 : 1

1 : 3

None of these

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Explanation

(b) Mass of satellite does not affects on orbital radius.

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The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will be

1/2 year

$2 \sqrt{2}$ years

4 years

8 years

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Explanation

(b) $\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2} = {(2)}^{3 / 2} = 2 \sqrt{2} \Rightarrow T_{2} = 2 \sqrt{2} y e a r s$

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The maximum and minimum distances of a comet from the sun are $8 \times 10^{12}$ m and $1.6 \times 10^{12}$ m. If its velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest -

112

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Explanation

(a) By conservation of angular momentum mvr = constant

$V_{m i n} \times V_{m a x} = V_{m a x} \times V_{m i n} ∴ V_{m i n} \frac{60 \times 1.6 \times 10^{12}}{8 \times 10^{12}} = \frac{60}{5} = 12 m / s$

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A body revolved around the sun 27 times faster then the earth . What is the ratio of their radii ?

1/3

1/9

1/27

1/4

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Explanation

(b)

$ω_{b o d y} = 27 ω_{e a r t h} T^{2} \propto r^{3} \Rightarrow ω \propto \frac{1}{r^{3 / 2}} \Rightarrow \frac{1}{r^{3 / 2}} ∴ r \propto \frac{1}{ω^{2 / 3}} \Rightarrow \frac{r_{b o d y}}{r_{e a r t h}} = {[\frac{ω_{e a r t h}}{ω_{b o d y}}]}^{2 / 3} = {(\frac{1}{27})}^{2 / 3} = \frac{1}{9}$

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The period of moon’s rotation around the earth is nearly 29 days. If moon’s mass were 2 fold, its present value and all other things remained unchanged, the period of moon’s rotation would be nearly -

$29 \sqrt{2} d a y s$

$29 / \sqrt{2} d a y s$

$29 \times 2 d a y s$

29 days

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Explanation

(d) Time period does not depends upon the mass of satellite.

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Two planets at mean distances $d_{1}$ and $d_{2}$ from the sun and their frequencies are n₁ and n₂ respectively, then;

${n_{1}}^{2} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$

${n_{2}}^{2} {d_{2}}^{3} = {n_{1}}^{2} {d_{1}}^{3}$

$n_{1} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$

${n_{1}}^{2} d_{1} = {n_{2}}^{2} d_{2}$

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Explanation

(b)

$\frac{T^{2}}{R^{3}} = \frac{T^{2}}{d^{3}} = \frac{1}{n^{2} d^{3}} = c o n s \tan t ∴ {n_{1}}^{2} {d_{1}}^{3} = {n_{2}}^{2} {d_{2}}^{2} [w h e r e n = f r e q u e n c y]$

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Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. , $F \propto 1 / r^{3}$ but still remaining a central force. Then -

Keplers law of areas still holds

Keplers law of period still holds

Keplers law of areas and period still hold

Neither the law of areas, nor the law of period still hold

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Explanation

(d) According to Keplers second law,

$\frac{∆ A}{∆ t} = \frac{L}{2 M}$

Since force is central, $τ = 0$

Hence L=constant

Now, $T α r^{3} / 2$

$F = \frac{G M m}{r^{3}} = \frac{m v^{2}}{r} \Rightarrow v = \frac{\sqrt{G M}}{r} T = \frac{2 {πr}^{2}}{r} \Rightarrow T α r^{2}$

So, Keplers law doesnot hold

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The distance of a planet from the sun is 5 times the distance between the earth and the sun. The time period of the planet is -

$5^{3 / 2}$ years

$5^{2 / 3}$ years

$5^{1 / 3}$ years

$5^{1 / 2}$ years

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Explanation

(a)

$\frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{{R_{1}}^{3}}{{R_{2}}^{3}} T_{2} = \sqrt{\frac{{R_{2}}^{3}}{{R_{1}}^{3}} {T_{1}}^{2}} = 5^{\frac{2}{3}} y e a r s$

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In an elliptical orbit under gravitational force, in general

Tangential velocity is constant

Angular velocity is constant

Radial velocity is constant

Areal velocity is constant

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Explanation

(d) Areal velocity = $\frac{d A}{d t} = \frac{L}{2 M}$

Here, L and M is cosntant

Areal velocity is constant

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