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a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Inside a small spaceship orbiting around the earth, the value of acceleration due to gravity can be calculated as constant and hence astronaut feels weightlessness.

If the space station orbiting around the earth has a large size such that variation in g matters in that case astronaut inside the spaceship will experience gravitational force and hence can detect gravity.

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- A body can be shielded from the gravitational influence of nearby matter because gravitational force between two point mass bodies is independent of the intervening medium. we cannot shield a body from gravitational influence of nearby matter.

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar
Explanation-
Yes, a body can have inertia but no weight. When we taken body to the centre of earth its has some inertia but no weight .

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Gravitational forces acting between two point masses m1 and m2, F= Gm1m2/r2 is independent of nature between them. So the gravitational force will remain unaffected when dipped in water.

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a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation-areal velocity of earth and sun

dA/dt=L/2m

where L is angular momentum

but we know L= r * P = r m v

dA/dt=1/2m (r * m v )=1/2 (r * v )

so dA/dt direction will be perpendicular to the plane containing r and v

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation-

areal velocity of a planet revolving around the sun is constant with respect to time.

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- examples of central force – gravitational force and electrostatic force

Examples of non central force = nuclear force, magnetic force acting between two current carrying loops etc.

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation-air molecules in the atmosphere are attracted vertically downward by the gravitational force of the earth just like an apple falling from the tree. Air molecules move randomly due to their thermal velocity and hence resultant motion of air molecules move randomly in vertical downward direction. But in case of apple it is heavier than air so it fall downwards only.

New answer posted

a year ago

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alok kumar singh

Contributor-Level 10

This is a Long Type Questions as classified in NCERT Exemplar

Explanation- rp= radius of perihelion =2R

Ra = radius of aphelion =6R

So ra=a(1+e)=6R

And rp= a(1-e)=2R

Solving above eqns

 We get e = ½

By conservation of angular momentum angular momentum at perigee= angular momentum at apogee

So mvprp=mvara

Va/vp=1/3

Where m is mass of satellite .

By conservation of energy , energy at perigee =energy at apogee

1 2 m v p 2 - G M m r p = 1 2 m v a 2 - G M m r a

So v p 2 1 - 1 9 = - 2 G M 1 r a - 1 r p = 2 G M 1 r p - 1 r a

Vp= [ 2 G M R ( 1 2 - 1 6 ) ] 1 / 2 [ [ 1 - v a v p ] 2 ] 1 / 2 = 3 4 G M R = 6.85 k m / s

Vp=6.85km/s , va=2.28km/s

So orbital velocity vc= G M r

R=6R ,vc= 3.23km/s

Hence to transfer to a circular orbit at apogee , we have to boost the velocity by 3.23-2.28=0.95km/s.

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

This is a Long Type Questions as classified in NCERT Exemplar

Explanation-let m be the mass of the earth vp, va be the velocity of the earth at perigee and apogee respectively. Similarly wp and wa are angular velocities.

At perigee, r p 2 w p = r a 2 w a at apogee

If a is the semimajor axis of the earth's orbit then rp=a (1-e) and ra=a (1+e)

w p w a = ( 1 + e 1 - e ) 2 e = 0.00167

w p w a = 1.0691

Let w be the angular speed which is geometric mean of wp and wa and corresponding to mean solar day.

w p w w w a =1.0691

If w corresponds to 10 per day then wp = 1.0340 per day and wa= 0.9670 per day . since 3610=24, mean solar day we get 361.0340 which corresponds to 24h,8.14' and 360.9670 corresponds to 23h59min52'. So

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