A 50 kg piece of space junk is falling towards Earth from a height of 1000 km above the Earth’s surface (mass of Earth is 6.0 ×1024kg, radius of Earth is 6.4 ×106m). Ignoring air resistance, determine:
a the gravitational field strength at this height
b the acceleration of the space junk at this height as it falls c the value of the ratio:
Solution
a At a height of 1000 km, the distance of the junk from the Earth’s centre is:
R=6.4 ×106+1.0 ×106=7.4 ×106m Therefore: g=
=
=7.3 N kg−1
b The acceleration of the space junk will equal the gravitational field strength, so it will fall with an acceleration of 7.3 m s−2towards Earth.
c You would expect the gravitational field to be stronger at the lower altitude. This can be confirmed by using a ratio approach:
At 500 km, R=6.9 ×106m At 1000 km, R=7.4 ×106m
= /
=
=
=1.2
The field strength at 500 km altitude is 1.2 times greater than the field strength at 1000 km.
(7.4 ×106) (6.9 ×106)2
R21000 km R2500 km
GM R21000 km
GM R2500 km
g(500 km) g(1000 km)
6.67 ×10−11×6.0 ×1024 (7.4 ×106)2 GM
R2
field strength at 500 km altitude field strength at 1000 km altitude
• A gravitational field is a region in which any body will experience a gravitational force.
• The gravitational field strength g is greater around more massive central objects, and becomes weaker at greater distances from the central body.
• Gravitational field strength g is given by:
g=
• At the surface of the Earth, the gravitational field strength is 9.8 N kg−1.
• The gravitational force acting on an object in a gravitational field is also called the weight, Fg, of the object and is given by:
Fg=mg=
• Any object that is falling freely through a gravitational field will fall with an acceleration equal to the gravitational field strength at that location.
GMm R2
GM R2
3.2 SUMMARY GRAVITATIONAL FIELDS
3.2 QUESTIONS
Assume that G=6.67 ×10−11N m2kg−2and that the gravitational field strength on the surface of the Earth, g, is 9.8 N kg−1.
1 Calculate the gravitational field strength, g, at the surface of each of the following planets, whose respective masses and radii are given in the following table:
Planet Mass Radius
Mercury 3.30 ×1023kg 2.44 ×106m Saturn 5.69 ×1026kg 6.03 ×107m Jupiter 1.90 ×1027kg 7.15 ×107m
2 Using your answers to question 1, calculate the weight of an 80 kg astronaut on the surface of each of the following planets:
a Mercury b Saturn c Jupiter.
3 The result of your calculation for question 1 should indicate that the gravitational field strength for Saturn is very close in value to that on Earth, i.e. approximately 10 N kg−1. However, the Earth’s radius and mass are very different from that of Saturn. How do you account for the fact that both planets have similar gravitational field strengths?
The following information applies to questions 4–6. A 100 kg meteor is falling towards the Earth from a distance of 4.0 Earth radii from the centre of the Earth (4.0RE).
4 Calculate the gravitational field strength at this height.
5 What is the acceleration of the meteor at this height?
6 For this meteor determine the ratio:
The following information applies to questions 7–9. There are bodies outside our Solar System, such as neutron stars, which produce very large gravitational fields. A typical neutron star may have a mass of 3.0 ×1030kg and a radius of 10 km.
7 Calculate the gravitational field strength at the surface of such a star.
8 Calculate the gravitational field strength at a distance of 5000 km from this star.
9 What would be the magnitude of the acceleration of a 10 000-tonne asteroid located at this distance and falling towards this star?
10 A gravimeter is a device that can measure the Earth’s gravitational field strength very accurately. Briefly explain how such a meter could be used to locate mineral deposits.
11 Two meteors, X and Y, are falling towards the Moon.
Both are 3.0 ×106m from the centre of the Moon.
Meteor X has a mass of 500 kg and meteor Y has a mass of 50 kg. The mass of the Moon is 7.3 ×1022kg.
Calculate:
a the gravitational force acting on X b the acceleration of X
c the gravitational force acting on Y d the acceleration of Y
e the gravitational field strength at this location.
12 There is a point between the Earth and the Moon where the total gravitational field is zero. The significance of this is that returning lunar missions are able to return to Earth under the influence of the Earth’s field once they pass this point. Given that the mass of Earth is 6.0 ×1024kg, the mass of the Moon is 7.3 ×1022kg and the radius of the Moon’s orbit is 3.8 ×108m, calculate the distance of this point from the centre of the Earth.
acceleration at 4.0RE acceleration at 2.0RE
A satellite is an object in a stable orbit around another object. Isaac Newton developed the notion of satellite motion while working on his theory of gravitation. He was comparing the motion of the Moon with the motion of a falling apple and realised that it was the gravitational force of attraction towards the Earth that determined the motion of both objects. He reasoned that if this force of gravity was not acting on the Moon, it would move with constant velocity in a straight line at a tangent to its orbit.
Newton proposed that the Moon, like the apple, was also falling. It was continuously falling to the Earth without actually getting any closer to the Earth. He devised a thought experiment in which he compared the motion of the Moon with the motion of a cannonball fired horizontally from the top of a high mountain.
3.3 Satellites in orbit
Figure 3.11 Newton realised that the gravitational attraction of the Earth was determining the motion of both the Moon and the apple.
(a)
(b)
Mars
Pluto Neptune
Jupiter Uranus Saturn
Earth
Sun
Mercury Venus
Mars Figure 3.13 The outer planets of the solar system have such large orbits that they cannot be shown on the same scale diagram as the inner planets. (a) The outer, or Jovian, planets are spread apart. If you travelled from the Sun to Pluto, by the time you reached Uranus you would be just halfway; (b) The inner, or terrestrial, planets are relatively small and close together. The asteroid belt lies between the orbits of Figure 3.12 Newton’s original sketch shows how a projectile that was fired fast enough would fall all the way around the Earth and become an Earth satellite.
In this thought experiment, if the cannonball was fired at a low speed it would not travel a great distance before gravitation pulled it to the ground. If it was fired with a greater velocity it would follow a less curved path and land a greater distance from the mountain. Newton reasoned that, if air resistance was ignored and if the cannonball was fired fast enough, it could travel around the Earth and reach the place from where it had been launched. At this speed it would continue to circle the Earth indefinitely. In reality satellites could not orbit the Earth at low altitudes, because of air resistance. Nevertheless Newton had proposed the notion of an artificial satellite almost 300 years before one was actually launched.
Fg
Fg