Tuesday, January 18, 2011

qUESTIONS OF GRAVITATION

Gravitation
Q.1 Define Gravitation. How is it different from gravity?
Q.2 State Newton’s law of gravitation. Express it in formvector
Q.3 What is “G” ? Mention its value, units and dimensions.
Q.4 Mention 5 important characteristics of gravitational force
Q.5 What is acceleration due to gravity? On what factors does it depend?
(b) What is its value at the earth’s surface?
Q.6 Assuming the earth to be a uniform sphere of radius 6400Kms and density 5.5gm/cc, find the value of g on the surface of earth
Q.7 Two metal spheres of mass M each are kept at a R. The gravitational force b/w them is found to be F. What will be the force b/w them if distance
a.         Mass of each is doubled while distance b/w them is made four times
b.         Mass of one of them is doubles while distance is kept R itself
c.         Both of them are kept in water at a distance R
Q.8 The mass of a planet is 1020 kg while its radius is 5000kms. What will be the gravitational force experienced by a spacecraft of mass 200kg its surface. standing on
(b) What is the acceleration near the surface of this planet
Q.9 The acceleration due to gravity near the earth’s surface is found to be 9.81m/s2. Radius of the earth is 6400 kms. Using this data find the mass of earth.
Q.10 Using the mass of earth calculated in the previous question, find the density of earth. In real life, is the density of earth uniform?
Q.11 Derive the value of acceleration due to gravity at a height h above the earth’s surface
Q.12 Derive the value of acceleration due to gravity at a depth d below the earth’s surface
Q.13 How does of earth the value of “g” ?affectthe shape
Q.14 How does the rotation of earth affect the value of “g” ? Derive the value of acceleration due to gravity at a place having λ.latitude
Q.15 With what angular velocity should the earth rotate so that a person at does not experience any gravity?the equator
Q.16 At what height above the earth’s surface does the value of “g” become one tenth of its value at the surface.
Q.17 At what height above the earth’s surface does the value of “g” become 64% of its value on the surface.
Q.18 Find the percentage change in of a body if it is taken to a height equal to the earth’s radiusthe weight
Q.19 At what depth below the earth’s surface does the value of “g” become 50% of its value on the surface.
Q.20 Find the value of acceleration due to gravity at the top of Mount Everest ( height 10 km )
Q.21 Determine the speed with which the earth should rotate so that a person on the equator would weigh 1/5th as much as present.
Q.22 Define Gravitational field. Mention its SI unit.
Q.23 What is gravitational potential? Mention its units and dimensions. Derive the value of gravitational potential at a distance r from a planet of mass m.
Q.24 Give 4 differences b/w gravitational and inertial mass.
Q.25 What is a satellite? Define geostationary satellite. What are the conditions required for it?
Q.26 A satellite of mass “m” is revolving about a planet of mass “M” in an orbit of radius “r” . Derive the expression for orbital velocity, time period and height of satellite.
Q.27 State Kepler’s laws of planetary motion. Derive the second and the third.
Q.28 The time period of Jupiter is 11.6 years. How far is Jupiter from the sun? Distance b/w earth and sun is known to be 1.5x10 11m.
Q.29 what is escape velocity? Find the value of escape velocity from earth.
(b) Find the value of escape velocity from a planet whose mass if double while radius is 4 times that of earth
Q.30 Two particles of mass 5kg and 10kg are kept 100m apart. Due to their mutual force of attraction they start approaching each other. Find the velocity of each of them when the separation b/w them has reduced to 25m.
(b) At what location will they collide?

PHYSICS'S PROBLEM

Gravitational Field Intensity and Universal Law of Gravity
G = 6.67 x 10-11 Nm2/kg2

Problems


1. A mass of 20.0 kg is located 4.0 m to the right of a mass of 30.0 kg.
a. What is the force on the 20.0 kg mass?
b. What is the force on the 30.0 kg mass?

2. At what point between the Earth and the Moon is the gravitational pull of the Earth equal in magnitude to that of the moon?

3. Find the altitude above the Earth's surface where Earth's gravitational field strength would be two-thirds of its value at the surface.

4.
Four masses form the vertices of a 1 m X 1 m square as shown. What is the gravitational force on the 3 kg mass due to the other three masses?







5. What force does Earth exert on a 80.0kg astronaut at an altitude equivalent to 2.5 times Earth's radius?


7. An object projected at 127 m/s upward from a small moon reaches a height of 5.08 km. If the radius of the moon is 1820 km, calculate its mass.

8. Estimate the surface gravity on a star that has five times the mass of our sun, and a radius of 10 km.

9. If the mass of Titan is 1.35*1023 kg and it's radius is 2570 km determine
a. the force at the surface on a 500 kg boulder.
b. the acceleration due to gravity.

10. What force does Earth exert on a 80.0kg astronaut at an altitude equivalent to 2.5 times Earth's radius?

11. What is the acceleration due to gravity at an altitude of 2*106 m above the earth\'s surface? Earth's mass is 5.98*1024 kg and of Earth's radius is 6.37*106 m .

12. A 492kg uniform solid sphere has a radius of .405m. Find the magnitude of the gravitational force (in N) exerted by the sphere on a 50.4g particle located 0.197m from the center of the sphere.

13. A hypothetical planet has a radius 2.0 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?

 Problems

(See below for answers)

1. satellite altitude
The period of Earth's moon is 27.3 d and has a mean orbital radius of 3.8 x 105 km. The Earth's radius is 6380 km. What is the altitude of a satellite orbiting the Earth once every 14.0 days? (search term: satellite altitude)

2. A satellite has a mass of 5850 kg and is in a circular orbit 4.1 * 105 m above the surface of a planet.

The period of the orbit is two hours. The radius of the planet is 4.15 * 106 m. What is the true weight of

the satellite when it is at rest on the planet's surface?

3. The orbit of the earth about the sun is almost circular. The closest and farthest distance are 1.47*108 and 1.52*108 km, respectively. Find the maximum variation in each of the following that result from the changing earth-sun distance in the course of 1 year.
a. kinetic energy
b. potential energy
c. total energy
d. orbital speed

4. At what horizontal velocity would a satellite have to be launched from the top of Mt. Everest to be placed in a circular orbit around the Earth?

5. The rings of a Saturn-like planet are composed of chunks of ice that orbit the planet. The inner radius of the rings is 78,000 km, while the outer radius is 190,000 km.The mass of this planet is 6.14*1026 kg. Find the period of an orbiting chunk of ice at the inner radius.

6. Calculate the kinetic energy of a 1249 kg Earth satellite in a circular orbit with a radius of 13740 miles.

7. Neptune is an average distance of 4.5*109 km from the sun. Estimate the length of the Neptunian year gives that the Earth is 1.50 * 108 km from the sun on the average?

8. What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 169km?
(a) in m/s
(b) What is the period of revolution?

9. Use the known period 27.3 days for the moon’s orbital motion around the earth. Given the radius of the orbit as 3.84x108m, calculate
a) the radius of the orbit of an earth satellite in a geosynchronous orbit with a period of 24 hours.
b)the mass of the earth.

10. Assume that you are agile enough to run across a horizontal surface at 8.5 m/s, independently of the value of the gravitational field. What would be the radius and the mass of an airless spherical asteroid of uniform density 1.1 x 103 kg/m3 on which you could launch yourself into orbit by running? What would be your period?

11. For a certain satellite with an apogee distance of ra=1.81*107m, the ratio of the orbital speed at perigee to the orbital speed at apogee is 1.11. Find the perigee distance rp, in meters.

12. Acceleration for a satellite is greater at its
a. same at both places b. perigee c. apogee

13. A satellite of mass 200 kg is launched from a site on Earth's equator into an orbit 200 km above the surface of Earth. Assuming a circular orbit and no air friction,
a. what is the orbital period of this satellite
b. what is the speed of the satellite in its orbit
c. what is the minimum energy necessary to place the satellite in orbit?

Sunday, January 16, 2011

E=mc


E = mc2 Explained

Albert Einstein
is perhaps the most famous scientist of this century. One of his most well-known accomplishments is the formula
Despite its familiarity, many people don't really understand what it means. We hope this explanation will help!
One of Einstein's great insights was to realize that matter and energy are really different forms of the same thing. Matter can be turned into energy, and energy into matter.
For example, consider a simple hydrogen atom, basically composed of a single proton. This subatomic particle has a mass of
0.000 000 000 000 000 000 000 000 001 672 kg
This is a tiny mass indeed. But in everyday quantities of matter there are a lot of atoms! For instance, in one kilogram of pure water, the mass of hydrogen atoms amounts to just slightly more than 111 grams, or 0.111 kg.
Einstein's formula tells us the amount of energy this mass would be equivalent to, if it were all suddenly turned into energy. It says that to find the energy, you multiply the mass by the square of the speed of light, this number being 300,000,000 meters per second (a very large number):

= 0.111 x 300,000,000 x 300,000,000
= 10,000,000,000,000,000 Joules
This is an incredible amount of energy! A Joule is not a large unit of energy ... one Joule is about the energy released when you drop a textbook to the floor. But the amount of energy in 30 grams of hydrogen atoms is equivalent to burning hundreds of thousands of gallons of gasoline!


If you consider all the energy in the full kilogram of water, which also contains oxygen atoms, the total energy equivalent is close to 10 million gallons of gasoline!
Can all this energy really be released? Has it ever been?

The only way for ALL this energy to be released is for the kilogram of water to be totally annhilated. This process involves the complete destruction of matter, and occurs only when that matter meets an equal amount of antimatter ... a substance composed of mass with a negative charge. Antimatter does exist; it is observable as single subatomic particles in radioactive decay, and has been created in the laboratory. But it is rather short-lived (!), since it annihilates itself and an equal quantity of ordinary matter as soon as it encounters anything. For this reason, it has not yet been made in measurable quantities, so our kilogram of water can't be turned into energy by mixing it with 'antiwater'. At least, not yet.

Another phenomenon peculiar to small elementary particles like protons is that they combine. A single proton forms the nucleus of a hydrogen atom. Two protons are found in the nucleus of a helium atom. This is how the elements are formed ... all the way up to the heaviest naturally occuring substance, uranium, which has 92 protons in its nucleus.
It is possible to make two free protons (Hydrogen nuclei) come together to make the beginnings of a helium nucleus. This requires that the protons be hurled at each other at a very high speed. This process occurs in
the sun, but can also be replicated on earth with lasers, magnets, or in the center of an atomic bomb. The process is called nuclear fusion.
What makes it interesting is that when the two protons are forced to combine, they don't need as much of their energy (or mass). Two protons stuck together have less mass than two single separate protons!
When the protons are forced together, this extra mass is released ... as energy! Typically this amounts to about 0.7% of the total mass, converted to an amount of energy predictable using the formula .

Elements heavier than iron are unstable. Some of them are very unstable! This means that their nuclei, composed of many positively charged protons, which want to repel from each other, are liable to fall apart at any moment! We call atoms like this radioactive.
Uranium, for example, is radioactive. Every second, many of the atoms in a chunk of uranium are falling apart. When this happens, the pieces, which are now new elements (with fewer protons) are LESS massive in total than the original uranium atoms. The extra mass disappears as energy ... again according to the formula ! This process is called nuclear fission.

Both these nuclear reactions release a small portion of the mass involved as energy. Large amounts of energy! You are probably more familiar with their uses. Nuclear fusion is what powers a modern nuclear warhead. Nuclear fission (less powerful) is what happens in an
atomic bomb (like the ones used against Japan in WWII), or in a nuclear power plant.

Albert Einstein was able to see where an understanding of this formula would lead. Although peaceful by nature and politics, he helped write a letter to the President of the United States, urging him to fund research into the development of an atomic bomb ... before the Nazis or Japan developed their own first. The result was the Manhatten Project, which did in fact produce the first tangible evidence of ... the atomic bomb!
http://www.e=mc2.com/

SUMIT

Hi I'm Sumit kumar sah.I studied at National Academy Campus.

Saturday, January 15, 2011