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Chapter 7: Problem 149

Mars is roughly 60 million km from Earth. How long does it take for a radio signal originating from Earth to reach Mars?

Short Answer

Expert verified
It takes 200 seconds for a radio signal originating from Earth to reach Mars, based on the given distance of 60 million km and the speed of light \( (3 \times 10^5\) km/s).

Step by step solution

01

Given Information

First, let's write down the information given in the problem: Distance from Earth to Mars = 60 million km Speed of radio signals (speed of light, c) = 300,000 km/s We need to find the time taken for a radio signal to cover this distance at the said speed.
02

Set up the equation

The formula for calculating the time taken to travel a certain distance is: time = distance / speed In this case, we need to find the time taken to travel the distance between Earth and Mars, which is 60 million km, at the speed of light, which is approximately 300,000 km/s. So the equation becomes: time = (60 million km) / (300,000 km/s)
03

Solve the equation

Now, let's solve the equation by plugging in the given values: time = (60 x 10^6 km) / (300,000 km/s) time = 200 s So, it takes 200 seconds for a radio signal originating from Earth to reach Mars.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Light
The speed of light is one of the fundamental constants in physics and is critically important in various calculations involving space and time. Light travels at an incredible speed of approximately 300,000 kilometers per second (km/s) in a vacuum. This speed is denoted by the letter "c" and is universally acknowledged for its magnitude and unchanging nature.
  • Light can travel around the Earth approximately 7.5 times in one second due to this remarkable speed.
  • This constant speed applies to all electromagnetic waves, including radio waves.
Knowing the speed of light enables us to perform precise calculations in physics, such as determining time lags for signals sent across vast distances. The concept of the speed of light allows scientists to better understand distances in space and the time it takes for information to travel.
Radio Signal Transmission
Radio signal transmission is a key method of communication across vast distances, such as between Earth and Mars. When a signal is transmitted from an antenna on Earth, it travels as an electromagnetic wave at the speed of light.
  • Radio waves are a type of electromagnetic radiation, similar to visible light but with longer wavelengths.
  • The transmission speed of these signals is also determined by the speed of light.
The ability to transmit signals at the speed of light means that the delay in receiving a message from Mars depends purely on the distance between the two planets. This understanding helps in planning communication tasks, such as sending commands to rovers or receiving data from space missions.
Distance Calculation
Calculating the distance between two points, such as Earth and Mars, is essential for understanding the time it takes signals to travel. Distance is often measured in kilometers when discussing astronomical distances. In this exercise, we know:
  • The distance from Earth to Mars is approximately 60 million kilometers.
Distance is a straightforward yet crucial variable in solving problems where time and speed are involved. Accurate distance measurements are essential in ensuring precise calculations for signal transmission timing and other space-related computations.
The relationship between distance, speed, and time is widely used in physics to determine any of these variables when the other two are known.
Time Calculation
Time calculation in this scenario involves determining how long it takes a radio signal to travel from Earth to Mars. By using the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \), we can solve for the time variable easily when distance and speed are known.
  • Substitute the known values: distance = 60 million km, speed = 300,000 km/s.
  • Perform the division: \( \text{time} = \frac{60 \times 10^6 \text{ km}}{300,000 \text{ km/s}} \).
  • The result is 200 seconds.
By understanding time calculation with this formula, students can apply this knowledge to similar physics problems involving movement across distances at constant speeds. Recognizing the role of speed and distance in determining travel time enhances problem-solving skills and provides a solid foundation for more advanced physics topics.

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Most popular questions from this chapter

The work function of an element is the energy required to remove an electron from the surface of the solid element. The work function for lithium is 279.7 kJ/mol (that is, it takes 279.7 kJ of energy to remove one mole of electrons from one mole of Li atoms on the surface of Li metal). What is the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal?

Write the expected ground-state electron configuration for the following: a. the element with one unpaired 5\(p\) electron that forms a covalent with compound fluorine b. the (as yet undiscovered) alkaline earth metal after radium c. the noble gas with electrons occupying 4f orbitals d. the first-row transition metal with the most unpaired electrons

Assume that a hydrogen atom's electron has been excited to the \(n=5\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of 253.4 \(\mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.)

For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}$$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{m}\right) .\) Calculate the position of the nodes for this wave function.
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