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Chapter 6: Problem 752

The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(\mathrm{R}=6.96 \times 10^{8} \mathrm{~m}\). The escape velocity of rocket from the sun \(\mathrm{is}=\ldots \ldots \ldots \mathrm{km} / \mathrm{sec}\) $\begin{array}{llll}\text { (A } 11.2 & \text { (B) } 12.38 & \text { (C) } 59.5 & \text { (D) } 618\end{array}$

Short Answer

Expert verified
The escape velocity of a rocket from the sun is approximately 617.542 km/sec, which is closest to answer choice D.

Step by step solution

01

The formula for escape velocity is given by \(v_{e} = \sqrt{ \frac{2GM}{R} }\), where \(v_{e}\) is the escape velocity, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2}\)), \(M\) is the mass of the celestial body (in this case, the sun), and \(R\) is the radius of the celestial body. #Step 2: Plug in the given values#

We are given the mass of the sun as \(1.99 \times 10^{30} \mathrm{kg}\) and the radius of the sun as \(6.96 \times 10^{8} \mathrm{m}\). We have the gravitational constant \(G = 6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2}\). Plug these values into the escape velocity formula: \(v_{e} = \sqrt{ \frac{2 \times 6.674 \times 10^{-11} \mathrm{m^3/kg \cdot s^2} \times 1.99 \times 10^{30} \mathrm{kg}}{6.96 \times 10^{8} \mathrm{m}} }\). #Step 3: Calculate the escape velocity in m/sec#
02

Simplify and calculate the escape velocity: \(v_{e} = \sqrt{ \frac{2 \times 6.674 \times 10^{-11} \times 1.99 \times 10^{30}}{6.96 \times 10^{8}} }\) \( \mathrm{m/sec} \approx 617,542 \mathrm{m/sec}\). #Step 4: Convert the escape velocity to km/sec#

To convert the escape velocity from m/sec to km/sec, divide the result by 1000: \(617,542 \mathrm{m/sec} \div 1000 = 617.542 \mathrm{km/sec}\). The escape velocity of a rocket from the sun is approximately 617.542 km/sec, which is closest to answer choice D.

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Most popular questions from this chapter

The escape velocity of a body from earth's surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth's surface will be (A) \((\mathrm{Ve} / \sqrt{2})\) (B) \((\mathrm{Ve} / 2)\) (C) \((\mathrm{Ve} / 2 \sqrt{2})\) (D) \((\mathrm{Ve} / 4)\)

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