Photo AI

Sir Isaac Newton deduced that the weight of an object is due to the force of gravity - Leaving Cert Physics - Question 6 - 2014

Question icon

Question 6

Sir-Isaac-Newton-deduced-that-the-weight-of-an-object-is-due-to-the-force-of-gravity-Leaving Cert Physics-Question 6-2014.png

Sir Isaac Newton deduced that the weight of an object is due to the force of gravity. Define force and give the unit of force. State Newton's law of universal gravi... show full transcript

Worked Solution & Example Answer:Sir Isaac Newton deduced that the weight of an object is due to the force of gravity - Leaving Cert Physics - Question 6 - 2014

Step 1

Define force and give the unit of force.

96%

114 rated

Answer

Force is defined as the cause of a change in motion. It is represented by the equation:

F=maF = ma

where:

  • FF is the force,
  • mm is the mass,
  • aa is the acceleration.

The unit of force is the Newton (N), which is defined as the force required to accelerate a mass of 1 kg at a rate of 1 m/s².

Step 2

State Newton's law of universal gravitation.

99%

104 rated

Answer

Newton's law of universal gravitation states that:

  • The force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

This can be expressed mathematically as:

Fm1m2r2F \propto \frac{m_1 m_2}{r^2}

Step 3

Use the equation below, which is from page 56 of the Formulae and Tables booklet, to calculate, to one decimal place, the acceleration due to gravity on Mars.

96%

101 rated

Answer

To find the acceleration due to gravity on Mars, we use the given formula:

g=GMr2g = \frac{GM}{r^2}

Where:

  • G=6.67×1011 N m2/kg2G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 (gravitational constant)
  • M=6.4×1023 kgM = 6.4 \times 10^{23} \text{ kg} (mass of Mars)
  • r=3.4×106 mr = 3.4 \times 10^6 \text{ m} (radius of Mars)

Now substituting values:

g=(6.67×1011)(6.4×1023)(3.4×106)2g = \frac{(6.67 \times 10^{-11}) (6.4 \times 10^{23})}{(3.4 \times 10^6)^2}

Calculating this gives:

g ≈ 3.7 m/s² (to one decimal place).

Step 4

Give a reason for the strength of the rover's wheels.

98%

120 rated

Answer

The wheels of the rover are not as strong as those of the vehicle that would be needed for Earth because Mars has a lower gravitational pull. The mass of the Mars rover, which is considerably lesser than that of Earth vehicles, and the less force exerted on the wheels results in the need for less robust construction.

Step 5

i) the weight of Curiosity on Earth.

97%

117 rated

Answer

The weight of an object can be calculated using the formula:

W=mgW = mg

Where:

  • m=899 kgm = 899 \text{ kg} (mass of Curiosity)
  • g=9.8 m/s2g = 9.8 \text{ m/s}^2 (acceleration due to gravity on Earth)

Thus:

W=899×9.88810.2 NW = 899 \times 9.8 ≈ 8810.2 \text{ N}

Therefore, the weight of Curiosity on Earth is approximately 8810 N.

Step 6

ii) the mass of Curiosity on Mars.

97%

121 rated

Answer

The mass of an object remains the same regardless of the gravitational field it is in. Thus, the mass of Curiosity on Mars is still:

899 kg.

Step 7

iii) the weight of Curiosity on Mars.

96%

114 rated

Answer

To find the weight on Mars, we again use the weight formula:

W=mgW = mg

Where:

  • m=899 kgm = 899 \text{ kg}
  • g=3.7 m/s2g = 3.7 \text{ m/s}^2 (acceleration due to gravity on Mars)

Thus:

W=899×3.73326.3 NW = 899 \times 3.7 ≈ 3326.3 \text{ N}

So, the weight of Curiosity on Mars is approximately 3326 N.

Step 8

The Curiosity rover communicates with Earth using radio waves, which are part of the electromagnetic spectrum.

99%

104 rated

Answer

Radio waves are part of the electromagnetic spectrum, which also includes microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. These waves share the characteristic of traveling at the speed of light in a vacuum, allowing the Curiosity rover to send and receive signals across vast distances.

Join the Leaving Cert students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;