Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?
Solution and Explanation
(a) No, gravitational force is a long range force and we cannot escape from it.
(b) Yes, Astronauts are unable to sense gravity because of the small spacecraft's weak gravitational pull from Earth. It is possible for him to detect the gravity if the space station is very huge, as this will also cause its magnitude to become noticeable.
(c) Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g). Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.
Top Questions on Gravitation
- At a height \( h \) above the Earth's surface, the acceleration due to gravity becomes \( \frac{g}{\sqrt{3}} \). What is the value of \( h \) in terms of the Earth's radius \( R \)?
- MHT CET - 2025
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- The value of \( g \) at height \( h \) above Earth's surface is \( \frac{g}{\sqrt{3}} \). Find \( h \) in terms of the radius of the Earth.
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- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R):
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
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A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
- JEE Main - 2025
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- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:
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Questions Asked in CBSE Class XI exam
- Consider the reactions:
- H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) \(\rightarrow\) H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
- H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) \(\rightarrow\) H3PO4(aq) + 2Cu(s) + H2SO4(aq)
- C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH- (aq) \(\rightarrow\) C6H5COO- (aq) + 2Ag(s) + 4NH3(aq) + 2 H2O(l)
- C6H5CHO(l) + 2Cu2+(aq) + 5OH-(aq) \(\rightarrow\) No change observed.
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
- CBSE Class XI
- Redox Reactions In Terms Of Electron Transfer Reactions
- Find \((x + 1)^6 - (x - 1)^6\). Hence or otherwise evaluate \((\sqrt2 + 1)^6 + (\sqrt2 - 1)^6.\)
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- Binomial Theorem for Positive Integral Indices
- If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}- CBSE Class XI
- Complement of a Set
- Consider the following species: N3- , O2-, F-, Na+, Mg2+ and Al3+
- What is common in them?
- Arrange them in the order of increasing ionic radii.
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- Modern Periodic Law And The Present Form Of The Periodic Table
- Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers.
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- How To Analyse Chemical Composition?
Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]