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Chapter 6: Problem 753

The mass of a space ship is \(1000 \mathrm{~kg} .\) It is to be launched from earth's surface out into free space the value of \(\mathrm{g}\) and \(\mathrm{R}\) (radius of earth) are \(10 \mathrm{~ms}^{-2}\) and \(6400 \mathrm{~km}\) respectively the required energy for this work will be $=\ldots \ldots \ldots .$ J (A) \(6.4 \times 10^{11}\) (B) \(6.4 \times 10^{8}\) (C) \(6.4 \times 10^{9}\) (D) \(6.4 \times 10^{10}\)

Short Answer

Expert verified
The energy required to launch the spaceship from Earth's surface into free space is approximately (C) \(6.4 \times 10^9 J\).

Step by step solution

01

Convert Earth's radius to meters

Since we are given the Earth's radius in kilometers, we need to convert it to meters. We know that 1 km = 1000 m, so we can easily convert 6400 km to meters: R = 6400 km * 1000 m/km = 6,400,000 m
02

Calculate the Gravitational Potential Energy (GPE)

The formula for calculating the GPE is: GPE = - (G * M * m) / r Where: - GPE is the gravitational potential energy (in joules) - G is the universal gravitational constant (approximately \(6.674 \times 10^{-11} N m^2/kg^2\)) - M is the mass of Earth (approximately \(5.972 \times 10^{24} kg\)) - m is the mass of the spaceship (1000 kg) - r is the distance from the center of Earth (Earth's radius) We already know the values for m and r. Let's plug them into the formula: GPE = - (\(6.674 \times 10^{-11} N m^2/kg^2\) * \(5.972 \times 10^{24} kg\) * 1000 kg) / 6,400,000 m
03

Calculate the energy required to overcome the gravitational force (work)

Now that we've calculated the GPE, to find the energy required, we need to find the work done against the Earth's gravitational force. Since GPE is negative, we need to make the result positive, and also multiply the GPE by the mass of the spaceship to find the energy. Energy required = - GPE Energy required = (\(6.674 \times 10^{-11} N m^2/kg^2\)) * \(5.972 \times 10^{24} kg\) * 1000 kg) / 6,400,000 m Energy required \( \approx 6.4 \times 10^9 J\)
04

Choose the correct answer

Comparing our result with the given options, we find that the energy required to launch the spaceship is approximately: (C) \(6.4 \times 10^9 J\)

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Most popular questions from this chapter

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

Two small and heavy sphere, each of mass \(\mathrm{M}\), are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is (A) zero (B) \(-(\mathrm{GM} / \mathrm{r})\) (C) \(-[(2 \mathrm{GM}) / \mathrm{r}]\) (D) \(-[(4 \mathrm{GM}) / \mathrm{r}]\)

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) \((12 / \sqrt{3})\left(\mathrm{Gm} / \mathrm{L}^{2}\right)\)

When a particle is projected from the surface of earth, it mechanical energy and angular momentum about center of earth at all time is constant (i) A particle of mass \(\mathrm{m}\) is projected from the surface of earth with velocity \(\mathrm{V}_{0}\) at angle \(\theta\) with horizontal suppose \(\mathrm{h}\) be the maximum height of particle from surface of earth and \(\mathrm{v}\) its speed at that point them \(\mathrm{V}\) is (A) \(\mathrm{V}_{0} \cos \theta\) \((\mathrm{B})>\mathrm{V}_{0} \cos \theta\) (C) \(<\mathrm{V}_{0} \cos \theta\) (D) zero (ii) Maximum height h of the particle is $(\mathrm{A})=\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (B) $>\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ $(\mathrm{C})<\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (D) can be greater than or less than $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

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