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Chapter 6: Problem 735

The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\) (B) \(112 \mathrm{kms}^{-1}\) (C) \(11.2 \mathrm{~ms}^{-1}\) (D) \(1120 \mathrm{kms}^{-1}\)

Short Answer

Expert verified
The escape velocity of a projectile from the earth is approximately (A) \(11.2 \mathrm{kms}^{-1}\).

Step by step solution

01

Understand the Escape Velocity Formula

The escape velocity formula is given by: \(v_e = \sqrt{\frac{2GM}{r}}\) where \(v_e\) = escape velocity \(G\) = gravitational constant, approximately \(6.674 \times 10^{-11} \frac{N(m/kg)^2}{m^2}\) \(M\) = mass of the earth, approximately \(5.972 \times 10^{24} kg\) \(r\) = radius of the earth, approximately \(6.371 \times 10^6 m\)
02

Substitute values and calculating the escape velocity

Plug the values of gravitational constant (G), mass of the Earth (M), and radius of the Earth (r) into the escape velocity formula and solve for the escape velocity (\(v_e\)). \(v_e = \sqrt{\frac{2 * (6.674 * 10^{-11}) * (5.972 * 10^{24})}{(6.371 * 10^6)}}\)
03

Perform the mathematical operations

Calculate the expression inside the square root. \(\frac{2 * (6.674 * 10^{-11}) * (5.972 * 10^{24})}{(6.371 * 10^6)} = 1.419 * 10^{17}\)
04

Find the square root

Now solve for \(v_e\). \(v_e = \sqrt{1.419 * 10^{17}} = 11.2 * 10^3 m/s\)
05

Interpret the result

The calculated escape velocity is \(11.2 * 10^3 \frac{m}{s}\), which can be expressed as \(11.2 \frac{km}{s}\). Therefore, the correct answer is (A) \(11.2 \mathrm{kms}^{-1}\).

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Most popular questions from this chapter

3 particle each of mass \(\mathrm{m}\) are kept at vertices of an equilateral triangle of side \(L\). The gravitational field at center due to these particles is (A) zero (B) \(\left[(3 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (C) \(\left[(9 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (D) \((12 / \sqrt{3})\left(\mathrm{Gm} / \mathrm{L}^{2}\right)\)

According to keplar, the period of revolution of a planet ( \(\mathrm{T}\) ) and its mean distance from the sun (r) are related by the equation (A) \(\mathrm{T}^{3} \mathrm{r}^{3}=\) constant (B) \(\mathrm{T}^{2} \mathrm{r}^{-3}=\) constant (C) \(\mathrm{Tr}^{3}=\) constant (D) \(\mathrm{T}^{2} \mathrm{r}=\) constant

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)

The period of a satellite in circular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C)

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The value of acc. due to gravity \((\mathrm{g})\) does not depend upon mass of the body Reason: This follows from \(\mathrm{g}=\left[(\mathrm{GM}) / \mathrm{R}^{2}\right]\), where \(\mathrm{M}\) is mass of planet (earth) and \(\mathrm{R}\) is radius of planet (earth) (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D
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