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Let us consider that our galaxy consists...

Let us consider that our galaxy consists of `2.5xx10^(11)` stars each of one solar mass. How long will this star at a distance of `50,000` light year from the galastic entre take to complete one revolution? Take the diameter of the Milky way to be `10^5ly. G=6.67xx 10^(-11) Nm^(2) Kg^(-2). (1 ly= 9.46xx10^(15)m)`

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To solve the problem, we will follow these steps: ### Step 1: Calculate the mass of the galaxy in kilograms Given that the galaxy consists of \(2.5 \times 10^{11}\) stars, each with a mass equal to one solar mass (\(M_{\odot} = 2 \times 10^{30} \, \text{kg}\)), we can calculate the total mass of the galaxy. \[ M = 2.5 \times 10^{11} \times 2 \times 10^{30} \, \text{kg} = 5.0 \times 10^{41} \, \text{kg} \] ...
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Find the mass of the earth from the following data. The period of lunar orbit around the earth is 27.3 days and radius of the orbit 3.9 xx 10^(5) km. G = 6.67 xx 10^(-11) Nm^(-2) kg^(-2) .

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Calculate the height above the earth at which the geostationary satellite is orbiting the earth. Radius of earth = 6400km. Mass of earth = 6 xx 10^(24) kg. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

Two masses 800 kg and 600kg are at a distance 0.25 m apart. Calculate the magnitude of the gravitational intensity at a point distant 0.20 m from the 800 kg and 0.15 m from the 600 kg mass. G = 6.66 xx 10^(-11) Nm^(2) kg^(-2) .

Calculate the gravitational force of attraction between two bodies of masses 40 kg and 80 kg separated by a distance 15 m. Take G= 6.7 xx 10^(-11) N m^2 kg^(-2)

You are given the following data :g=9.81ms^(-2) , radius of earth =6.37xx10^(6)m the distance of the Moon from the earth =3.84xx10^(8) m and the time period of the Moon's revolution =27.3days . Obtain the mass of the earth in two different ways. G=6.67xx10^(-11)Nm^(2)kg^(-2) .

If the acceleration due to gravity on earth is 9.81 m//s^(2) and the radius of the earth is 6370 km find the mass of the earth ? (G = 6.67 xx 10^(-11) Nm^(2)//kg^(2))

Three masses of 1 kg, 2kg, and 3 kg are placed at the vertices of an equilateral triangle of side 1m. Find the gravitational potential energy of this system (Take, G = 6.67 xx 10^(-11) "N-m"^(-2)kg^(-2) )

Determine the force required to double the length of the steel wire of area of cross-section 5 xx 10^(-5) m^(2) . Give Y for steel = 2 xx 10^(11) Nm^(-2) .

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NCERT ENGLISH-GRAVITATION-EXERCISE
  1. Suppose there existed a planet that went around the sun twice as fast ...

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  2. One of the satellite of jupiter, has an orbital period of 1.769 days a...

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  3. Let us consider that our galaxy consists of 2.5xx10^(11) stars each of...

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  4. Choose the correct alternative : (a) If the zero of the potential en...

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  5. Does the escape speed of a body from the earth depend on (a) the mass ...

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  6. A comet orbits the Sun in a highly elliptical orbit. Does the comet ha...

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  7. Which of the following symptoms is likely to afflict an astronaut in s...

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  8. In the following two exercises, choose the correct answer from among t...

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  9. For the above problem, the direction of the gravitational intensity at...

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  10. A rocket is fired from the earth towards the sun. At what distance fro...

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  11. The mean orbital radius of the Earth around the Sun is 1.5 xx 10^8 km....

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  12. A Saturn year is 29.5 times that earth year. How far is the Saturn fro...

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  13. A body weighs 63 N on the surface of the earth. What is the gravitatio...

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  14. Assuming the earth to be a sphere of uniform mass density, how much wo...

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  15. A rocket is fired vertically with a speed of 5 kms^(-1) from the earth...

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  16. The escape speed of a projectile on the earth's surface is 11.2 km s^(...

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  17. A satellite orbits the earth at a height of 400 km, above the surface....

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  18. Two starts each of one solar mass (=2xx10^(30)kg) are approaching each...

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  19. Two heavy sphere each of mass 100 kg and radius 0.10 m are placed 1.0 ...

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  20. A geostationary satellite orbits the Earth at a height of nearly 36,00...

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