The mean radius of the earth’s orbit around the sun is 1.5 × 10 11 m. The mean radius of the orbit of mercury around the sun is 6 × 10 10 m. Calculate the year of the mercury.

Given in the question, Radius of the earth's orbit around the sun is, a E = 1 · 5 × 10 11 m Radius of the mercury, a m = 6 × 10 10 m Now for one year, From Kepler's third law, T m T E 2 = a m a E 3 ⇒ T m 2 1 y e a r = 6 × 10 10 1 · 5 × 10 11 3 = 2 5 3 ⇒ T m = 2 5 3 2 Year So, the year for mercury 2 5 3 / 2 y e a r s .

The time period of revolution of moon around the earth is 28 days and radius of its orbit is 4 × 10 5 km. If G = 6.67 × 10 –11 Nm 2 /kg 2 then find the mass of the earth.

Given in the question that, The time period of revolution of moon around the earth is, T = 28 × 24 × 60 × 60 s And the radius of its orbit is, a = 4 × 10 5 × 1000 m And the gravitational constant is, G = 6 · 67 × 10 - 11 Nm 2 / kg 2 Now, from Kepler's law, T 2 = 4 π 2 G M a 3 ⇒ M = 4 π 2 a 3 G T 2 ⇒ M = 4 × 3 · 14 × 3 · 14 × 4 × 10 8 3 6 · 67 × 10 - 11 × 28 × 24 × 3600 2 ⇒ M = 6 · 47 × 10 24 kg Hence, the mass of earth is 6 · 67 × 10 24 kg .

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The gravitational field due to a mass distribution is, $E = \frac{K}{x^{3}}$ in the $x$ -direction. Here, $K$ is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance $x$ is,

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Important Questions on Gravitation

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 3>Q 7

A body of mass

m

is situated at distance

4 R_{e}

above the earth's surface, where

R_{e}

is the radius of earth how much minimum energy be given to the body so that it may escape :-

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 1

The mean radius of the earth’s orbit around the sun is 1.5 × 10¹¹ m. The mean radius of the orbit of mercury around the sun is 6 × 10¹⁰ m. Calculate the year of the mercury.

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 2

If earth describes an orbit round the sun of double its present radius, what will be the year on earth ?

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 3

If the gravitational force were to vary inversely as

m^{th}

power of the distance, then the time period of a planet in a circular orbit of radius

r

around the sun will be proportional to:

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 4

A planet is revolving around the sun in an elliptical orbit. Its closest distance from the sun is

r_{\min}

. The farthest distance from the sun is

r_{\max}

. If the orbital angular velocity of the planet when it is nearest to the sun is

ω

, then the orbital angular velocity at the point when it is at the farthest distance from the sun is:-

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 5

The time period of revolution of moon around the earth is 28 days and radius of its orbit is 4 × 10⁵ km.
If G = 6.67 × 10^–11 Nm²/kg² then find the mass of the earth.

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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 6

Let the speed of the planet at the perihelion P in Fig. be v_P and the Sun-planet distance SP be r_p. Relate (r_p, v_P) to the corresponding quantities at the aphelion (r_A, v_A). Will the planet take equal times to traverse BAC and CPB ?
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Beta Question Bank for Medical: Physics>Chapter 11 - Gravitation>BEGINNER'S BOX - 4>Q 7

A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant linear speed, angular speed, angular momentum, kinetic energy, potential energy, total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

The gravitational field due to a mass distribution is, E=Kx3 in the x-direction. Here, K is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance x is,

Important Questions on Gravitation

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The gravitational field due to a mass distribution is, $E = \frac{K}{x^{3}}$ in the $x$ -direction. Here, $K$ is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance $x$ is,