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Chapter 7: Problem 68

The time period of an earth satellite in circular orbit is independent of (A) the mass of the satellite. (B) radius of its orbit. (C) both the mass and radius of the orbit. (D) neither the mass of the satellite nor the radius of its orbit.

Short Answer

Expert verified
The time period of an Earth satellite in a circular orbit is independent of (A) the mass of the satellite.

Step by step solution

01

Gravitational force acting on the satellite

The gravitational force acting on a satellite in orbit can be represented as: \(F_g = G \frac{M_Em_s}{r^2}\) Where: - \(F_g\) = gravitational force between the Earth and the satellite - \(G\) = gravitational constant - \(M_E\) = mass of the Earth - \(m_s\) = mass of the satellite - \(r\) = distance between the center of the Earth and the satellite (radius of the orbit)
02

Centripetal force acting on the satellite

The centripetal force required for the satellite to maintain its circular path is given by: \(F_c = m_s \frac{v^2}{r}\) Where: - \(F_c\) = centripetal force - \(v\) = orbital speed of the satellite
03

Equate gravitational force and centripetal force

Since the centripetal force needed by the satellite to remain in orbit is provided by the gravitational force, we can equate the two equations: \(G \frac{M_Em_s}{r^2} = m_s \frac{v^2}{r}\)
04

Solve for orbital speed

We can now solve for the orbital speed of the satellite: \(v = \sqrt{G\frac{M_E}{r}}\)
05

Derive the formula for the time period of a satellite

The time period (T) of the satellite is given by the ratio of the circumference of the orbit (2蟺r) to the orbital speed (v): \(T = \frac{2 \pi r}{v}\) Plugging the value of \(v\) from Step 4 into the equation: \(T = \frac{2 \pi r}{\sqrt{G\frac{M_E}{r}}}\)
06

Simplify the equation for the time period

We can now simplify the equation for the time period as: \(T = 2 \pi \sqrt{\frac{r^3}{GM_E}}\)
07

Analyze the given options

From the derived formula, we can see that the time period (T) of the satellite is independent of: - the mass of the satellite (m_s) -- Option (A) And depends on: - the radius of its orbit (r) -- Option (B) Therefore, the correct answer is: The time period of an Earth satellite in a circular orbit is independent of (A) the mass of the satellite.

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Most popular questions from this chapter

Three planets of same density have radii \(R_{1}, R_{2}\) and \(R_{3}\) such that \(R_{1}=2 R_{2}=3 R_{3}\). The gravitational field at their respective surfaces are \(g_{1}, g_{2}\) and \(g_{3}\) and escape velocities from their surfaces are \(v_{1}, v_{2}\) and \(v_{3}\) respectively, then (A) \(g_{1} / g_{2}=2\) (B) \(g_{1} / g_{3}=3\) (C) \(v_{1} / v_{2}=1 / 4\) (D) \(v_{1} / v_{3}=3\)

A particle initially at rest is displaced by applying a non-conservative force \(F\) in a uniform gravitational field. In the process, following physical quantities associated with the particle are measured. \(\Delta U=\) change in gravitational potential energy \(\Delta K=\) change in kinetic energy \(\Delta W_{1}=\) work done by the force \(F\) \(\Delta W_{2}=\) work done by the gravitational force (A) \(\Delta W_{2}=-\Delta U\) (B) \(\Delta K=\Delta W_{1}+\Delta W_{2}\) (C) \(\Delta K+\Delta U=\Delta W_{1}+\Delta W_{2}\) (D) \(\Delta W_{1}=\Delta W_{2}\)

Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth, This is because, (A) the solar cells and batteries in satellites run out. (B) the laws of gravitation predict a trajectory spiralling inwards. (C) of viscous forces causing the speed of satellite and hence height to gradually decrease. (D) of collisions with other satellites.

The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) (where \(g\) = the acceleration due to gravity on the surface of the earth) in terms of \(R\), the radius of the earth is (A) \(\frac{R}{\sqrt{2}}\) (B) \(R / 2\) (C) \(\sqrt{2} R\) (D) \(2 R\)

A simple pendulum has a time period \(T_{1}\) when on the earth's surface, and \(T_{2}\) when taken to a height \(R\) above the earth's surface, where \(R\) is radius of earth. The value of \(T_{2} / T_{1}\) is (A) 1 (B) \(\sqrt{2}\) (C) 4 (D) 2
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