Planet and Satellite

Practice 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,

1. the acceleration of S is always directed towards the centre of the earth.

2. the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.

3. the total mechanical energy of S varies periodically with time.

4. the linear momentum of S remains constant in magnitude.

View Answer

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

1. T is conserved

2. U is always positive

3. E is always negative

4. L is conserved but direction of vector L will continously change

View Answer

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

1. [M L T–³] and [M L T–4]

2. [M L T–4] and [M L T–³]

3. [M L T–¹] and [M L T–²]

4. [M L T–²] and [M L T0]

View Answer

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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