/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A satellite of mass \(\mathrm{m}... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 67

A satellite of mass \(\mathrm{m}\) revolves around the earth of radius \(R\) at a height \(x\) from its surface. If \(g\) is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is (A) \(g x\) (B) \(\frac{g R}{R-x}\) (C) \(\frac{g R^{2}}{R+x}\) (D) \(\left(\frac{g R^{2}}{R+x}\right)^{1 / 2}\)

Short Answer

Expert verified
The orbital speed of the satellite is \(v = \left(\frac{g R^{2}}{R+x}\right)^{1/2}\), so the correct choice is (D).

Step by step solution

01

Find the Gravitational Force

We will use Newton's law of gravitation to find the force between the Earth and the satellite. The gravitational force F between two objects with masses m1 and m2 separated by a distance r is given by: \(F = G \frac{m_{1} m_{2}}{r^{2}}\) where G is the gravitational constant. In our case, m1 is the mass of the Earth (M), m2 is the mass of the satellite (m), and the distance r between the centers of both objects is the sum of the Earth's radius R and the height x (i.e., r = R + x). Hence, the gravitational force between the Earth and the satellite is: \(F = G \frac{M m}{(R+x)^{2}}\)
02

Determine the Centripetal Force

Now we will find the centripetal force that keeps the satellite in orbit. The centripetal force Fc acting on an object of mass m moving in a circular path of radius r at a speed v is given by: \(F_{c} = m \frac{v^{2}}{r}\) In our case, the mass is that of the satellite (m) and the radius of the circular path is the sum of the Earth's radius R and the height x (i.e., r = R + x). So, the centripetal force acting on the satellite is: \(F_{c} = m \frac{v^{2}}{R + x}\)
03

Equate Gravitational Force and Centripetal Force

Since the gravitational force provides the centripetal force required to keep the satellite in orbit, we will now equate these two forces: \(G \frac{M m}{(R+x)^{2}} = m \frac{v^{2}}{R + x}\)
04

Solve for Orbital Speed (v)

Divide both sides of the equation by m and multiply both sides by (R + x): \(G M = v^{2} (R+x)\) To find the orbital speed v, take the square root of both sides: \(v = \sqrt{\frac{G M}{R+x}}\) Now, given that g is the acceleration due to gravity on the Earth's surface, we can write: \(g = \frac{G M}{R^2}\) So, GM can be expressed in terms of g and R: \(G M = g R^2\) Substitute GM in the formula for v: \(v = \sqrt{\frac{g R^2}{R+x}}\) Comparing this expression to the given options, we find that the correct answer is: \(v = \left(\frac{g R^{2}}{R+x}\right)^{1/2}\) So, the correct choice is (D).

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Most popular questions from this chapter

Different points in the earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the CM (centre of mass) causing translation and a net torque at the CM causing rotation around an axis through the CM. For the earth-sun system (approximating the earth as a uniform density sphere). (A) The torque is zero. (B) The torque cause the earth to spin. (C) The rigid body result is not applicable since the earth is not even approximately a rigid body. (D) The torque causes the earth to move around the sun.

Two objects of masses \(m\) and \(4 m\) are at rest at an infinite separation. They move towards each other under mutual gravitational force of attraction. If \(G\) is the universal gravitational constant. Then at separation \(r\) (A) the total energy of the two objects is zero. (B) their relative velocity of approach is \(\left(\frac{10 G m}{r}\right)^{\frac{1}{2}}\) in (C) the total kinetic energy of the objects is \(\frac{4 G m^{2}}{r}\). (D) net angular momentum of both the particles is zero about any point.

Two spherical bodies of mass \(M\) and \(5 M\) and radii \(R\) and \(2 R\) respectively, are released in free space with initial separation between their centres equal to \(12 R\). If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is (A) \(2.5 R\) (B) \(4.5 R\) (C) \(7.5 R\) (D) \(1.5 R\)

Suppose you are, as the header of a group of scientists working in NASA, sent to a planet named NSP2009 to study that planet you have following data with you. 1\. The planet NSP2009 is spherical in shape and of radius \(R\). 2\. The mass of the planet NSP2009 is M. If your team finds that the weight of any body inside the planet remains same as on its surface, what could be the possible value of mass density of the planet as a function of radial distance (where \(\rho_{0}\) is a content) \((R \geq>0)\) (A) \(\rho=\rho_{0} r^{0}\) (B) \(\rho=\rho_{0} r\) (C) \(\rho=\rho_{0} r^{-1}\) (D) \(\rho=\rho_{0} r^{-1 / 2}\)

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{~km} / \mathrm{s}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be (A) \(11 \sqrt{2} \mathrm{~km} / \mathrm{s}\) (B) \(22 \mathrm{~km} / \mathrm{s}\) (C) \(11 \mathrm{~km} / \mathrm{s}\) (D) \(\frac{11}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)
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