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Chapter 6: Problem 737

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

Short Answer

Expert verified
The correct formula for escape velocity is (C) \(U=\sqrt{2 \mathrm{GMR}}\).

Step by step solution

01

Establish mass, density, and volume relationship

To find the correct formula, we first need a relationship between mass, density, and volume. We know that mass (M) equals density (ρ) times volume (V). Therefore, we have: \[M = \rho V\] Since Earth can be approximated as a sphere, we'll use this to compute the volume of Earth: \[V = \frac{4}{3} \pi R^{3}\] where R is the radius of Earth.
02

Calculate gravitational force

The gravitational force (F) between two bodies with masses (M1 and M2) and a distance R separating them is given by: \[F = G\frac{M_{1}M_{2}}{R^{2}}\] In our case, M1 is the mass of Earth (M) and M2 is the mass of the object trying to escape Earth. We can rewrite the equation as: \[F = G\frac{Mm}{R^{2}}\]
03

Equate gravitational force to the work done

Escape velocity is defined as the minimum velocity an object needs to be propelled to escape the gravitational pull of a massive body, like Earth, without further propulsion. The work done against gravity must be equal to the gravitational force. Work done (W) can be given by: \[W = Fd\] where d is the distance traveled (in our case, it is equal to the radius R of Earth). We can rewrite the equation as: \[W = GMm / R\]
04

Determine kinetic energy and escape velocity

The work done (W) is equal to the kinetic energy (K.E.) of the object which is trying to escape Earth's gravity, given by: \[K.E. = \frac{1}{2}mv^{2}\] where m is the mass of the object and v is its velocity. Set the work done equal to the kinetic energy: \[GMm / R = \frac{1}{2}mv^{2}\] where v represents the escape velocity.
05

Solve for escape velocity

Next, we solve for v (escape velocity) from the equation derived in Step 4. First, divide both sides by the mass of the object (m): \[\frac{GM}{R} = \frac{1}{2}v^{2}\] Then, multiply both sides by 2 to eliminate the 1/2: \[2\frac{GM}{R} = v^{2}\] Finally, take the square root of both sides: \[v = \sqrt{2\frac{GM}{R}}\] Comparing this result to the given options, we find that the correct formula for escape velocity is: (C) \(U=\sqrt{2 \mathrm{GMR}}\)

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Most popular questions from this chapter

The acceleration of a body due to the attraction of the earth (radius R) at a distance \(2 \mathrm{R}\) from the surface of the earth is \(=\) (g \(=\overline{\text { acceleration due to gravity at the surface of earth })}\) (A) \(\mathrm{g} / 9\) (B) \(\mathrm{g} / 3\) (C) \(\mathrm{g} / 4\) (D) 9

Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be \(-m s^{-2}\) \(\left(\mathrm{g}=9.8 \mathrm{~ms}^{2}\right)\) (A) \(19.6\) (B) \(9.8\) (C) \(4.9\) (D) \(2.45\)

Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius \(\mathrm{R}\) around the sun will be proportional to (A) \(\mathrm{R}^{[(\mathrm{n}+1) / 2]}\) (B) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\) (C) \(\mathrm{R}^{\mathrm{n}}\) (D) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\)

A satellite is moving around the earth with speed \(\mathrm{v}\) in a circular orbit of radius \(\mathrm{r}\). If the orbit radius is decreased by \(1 \%\) its speed will (A) increase by \(1 \%\) (B) increase by \(0.5 \%\) (C) decreased by \(1 \%\) (C) Decreased by \(0.5 \%\)

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite
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