NEET Planet and Satellite MCQs & PYQs

Q: A planet is moving in a elliptical orbit. If T, U, E and L are its kinetic energy, potential energy, total energy and magnitude of angular momentum respectively. Then which is true

For a planet in an elliptical orbit: - Total energy (E) is the sum of kinetic energy (T) and potential energy (U). - In a bound orbit like an ellipse, E is always negative. This indicates that the planet is gravitationally bound to the star and cannot escape. - Kinetic energy (T) is always positive. - Potential energy (U) is negative due to the attractive gravitational force, and its magnitude is greater than T. - Angular momentum (L) is constant for elliptical orbits. Thus, E is always negative for elliptical orbits.

Practice NEET Planet and Satellite Questions

Question 1:

easy

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,

Yes, the acceleration of the satellite \( S \) is always directed towards the center of the Earth. This is because the gravitational force, which provides the acceleration, always points towards the Earth's center, regardless of the satellite's position in its elliptical orbit. This centripetal acceleration is responsible for keeping the satellite in orbit.

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Question 2:

easy

A planet is moving in a elliptical orbit. If T, U, E and L are its kinetic energy, potential energy, total energy and magnitude of angular momentum respectively. Then which is true

For a planet in an elliptical orbit:

- Total energy (E) is the sum of kinetic energy (T) and potential energy (U).
- In a bound orbit like an ellipse, E is always negative. This indicates that the planet is gravitationally bound to the star and cannot escape.
- Kinetic energy (T) is always positive.
- Potential energy (U) is negative due to the attractive gravitational force, and its magnitude is greater than T.
- Angular momentum (L) is constant for elliptical orbits.

Thus, E is always negative for elliptical orbits.

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Question 3:

easy

A force F is given by F = at + bt², where t is time. The dimensions of a and b are

The force \( F = at + bt^2 \) has dimensions of force \([M L T^{-2}]\).

For \( at \), the dimensions of \( a \) must be:

\[
[M L T^{-2}] = [a][T]
\]

Thus, the dimensions of \( a \) are:

\[
a = [M L T^{-3}]
\]

For \( bt^2 \), the dimensions of \( b \) must be:

\[
[M L T^{-2}] = [b][T^2]
\]

Thus, the dimensions of \( b \) are:

\[
b = [M L T^{-4}]
\]

So, the dimensions are:
- \( a = [M L T^{-3}] \)
- \( b = [M L T^{-4}] \)

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Question 4:

easy

The escape velocity of a body on the earth surface is 11.2 km/s. If the same body is projected upward with velocity 22.4 km/s, the velocity of this body at infinite distance from the centre of the earth will be

Using conservation of energy: \(v_\infty = \sqrt{v^2 - v_e^2}\). Given \(v = 22.4\text{ km/s} = 2v_e\), we find \(v_\infty = \sqrt{(2v_e)^2 - v_e^2} = v_e\sqrt{3} = 11.2\sqrt{3}\text{ km/s}\).

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Question 5:

easy

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy \( E_0 \). Its potential energy is:

For a satellite in circular orbit, Potential Energy is \( U = -\frac{GMm}{r} \) and Total Energy is \( E_0 = -\frac{GMm}{2r} \). Thus, Potential Energy is twice the Total Energy, \( U = 2E_0 \).

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Question 6:

easy

Assertion: If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases.

Reason: The speed of satellite is a constant quantity.

As a satellite moves to a lower orbit, total energy decreases (becomes more negative) due to dissipation, but kinetic energy increases, so speed increases. The speed of a satellite is not universally constant.

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Question 7:

easy

A satellite is seen after each 8 hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth & satellite is same will be :

When rotating oppositely, \(\frac{1}{T_{\text{rel}}} = \frac{1}{T_s} + \frac{1}{T_e} \Rightarrow \frac{1}{8} = \frac{1}{T_s} + \frac{1}{24}\), which gives \(T_s = 12\text{ hours}\). When rotating in the same direction, \(\frac{1}{T_{\text{rel}}'} = \frac{1}{T_s} - \frac{1}{T_e} = \frac{1}{12} - \frac{1}{24} = \frac{1}{24}\), so \(T_{\text{rel}}' = 24\text{ hours}\).

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Question 8:

easy

A satellite of mass \(m\) is in a circular orbit of radius \(2R\) about the earth. How much energy is required to transfer it to a circular orbit of radius \(4R\) : (\(R =\) Radius of earth)

The total energy of a satellite is \(E = -\frac{GMm}{2r}\). The required energy is \(\Delta E = E_f - E_i = -\frac{GMm}{8R} - \left(-\frac{GMm}{4R}\right) = \frac{GMm}{8R}\). Since \(g = \frac{GM}{R^2}\), we get \(\Delta E = \frac{mgR}{8}\).

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Question 9:

easy

If the earth be at one half its present distance from the sun, number of days in the year will be nearly

According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).

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Question 10:

easy

The period of a satellite in a circular orbit of radius \(R\) is \(T\). What is the period of another satellite in a circular orbit of radius \(4R\) ?

By Kepler's Third Law, \(T^2 \propto R^3\). Therefore, \(\frac{T'}{T} = \left(\frac{4R}{R}\right)^{3/2} = 8\), which gives \(T' = 8T\).

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