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Chapter 8: Problem 39

What is the value of escape velocity on the earth?

Short Answer

Expert verified
Answer: The escape velocity for Earth is approximately 11,186 m/s.

Step by step solution

01

Gather the necessary information

Before we calculate the escape velocity, we need some values. In this case, the mass of the Earth (M) and the radius of the Earth (R). The known values are: Mass of the Earth (M) = 5.972 x 10^24 kg Radius of the Earth (R) = 6.371 x 10^6 m
02

Understand the escape velocity formula

The formula for calculating escape velocity is: \(v_{escape} = \sqrt{\frac{2GM}{R}}\) Where G is the gravitational constant (6.67430 × 10^{-11} m^3 kg^{-1} s^{-2}), M is the mass of the Earth, R is the radius of the Earth, and \(v_{escape}\) is the escape velocity we want to calculate.
03

Plug in the values and solve the formula

Now we have all the values we need to calculate the escape velocity. Simply plug them into the formula and solve: \(v_{escape} = \sqrt{\frac{2 \times 6.67430 × 10^{-11} m^3 kg^{-1} s^{-2} \times 5.972 × 10^{24} kg}{6.371 × 10^6 m}}\)
04

Calculate the escape velocity

Perform the calculation, making sure to follow the order of operations (PEMDAS): \(v_{escape} = \sqrt{\frac{2 × 6.67430 × 10^{-11} m^3 kg^{-1} s^{-2} \times 5.972 × 10^{24} kg}{6.371 × 10^6 m}} = 11186.4 m/s\) The escape velocity for Earth is approximately 11,186 m/s.

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Most popular questions from this chapter

The time interval between two successive primary and secondary tides at a place is more than 12 hours because (1) of the rotation of earth about its axis (2) the moon also moves round the earth in the same direction of the earth's rotation (3) the moon also moves round the earth in the opposite direction of the earth's rotation (4) the earth revolves round the sun

A barometer can be used to determine the (1) temperature (2) humidity (3) altitude (4) None of these

Name the device used to find the atmospheric pressure.

Two bodies of masses \(\mathrm{m}\) and \(2 \mathrm{~m}\) are separated by a distance 'd'. The body of mass \(\mathrm{m}\) attracts another body of mass \(2 \mathrm{~m}\) with a force \(\mathrm{F}\), then the body of mass \(2 \mathrm{~m}\) attracts \(\mathrm{m}\) with the force of (1) \(\mathrm{F}\) (2) \(2 \mathrm{~F}\) (3) \(\frac{\mathrm{F}}{2}\) (4) \(3 \mathrm{~F}\)

The mass of a planet is double the mass of the earth and its radius is half of that of the earth. If \(\mathrm{F}\) is the force of attraction on an object on the surface of earth, then the force of attraction on the same object on the surface of that planet is (1) \(2 \mathrm{~F}\) (2) \(4 \mathrm{~F}\) (3) \(\underline{\mathrm{F}}\) (4) \(8 \mathrm{~F}\)
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