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1 | 2 | 3 |
---|---|---|

mass | m/s^{2} |
zero at the center |

weight | kg | measure of inertia |

acceleration due to gravity | Nm^{2}/kg^{2} |
same in the entire universe |

gravitational constant | N | depends on height |

Answer ::

1 | 2 | 3 |
---|---|---|

mass | N | measure of inertia |

weight | kg | zero at the center |

acceleration due to gravity | m/s^{2} |
depends on height |

gravitational constant | Nm^{2}/kg^{2} |
same in entire universe |

Answer::

Mass is the amount of matter present in an organism while weight of an object is mass along with gravitational force I. e. W=Mg.

Mass of an object will be same at any point of the universe but its weight depends on the amount of gravity present on that planet thus mass of object on Mars will be same but weight will be different. {*this question can come for 2 marks in exam *}

Answer::

Whenever an object moves under the influence of the force of gravity alone, it is said to be in Free Fall. During free fall initial velocity of an object is zero and force of air also acts on an object. Thus real free fall is possible only in a vacuum because there is no air.

The earth exerts gravitational force on objects near it. According to Newton’s second law of motion, a force acting on a body results in its acceleration. Thus, the gravitational force due to the earth on a body results in its acceleration. This is called acceleration due to gravity and is denoted by ‘g’.

The minimum value of the initial velocity at which an object to escape from the gravitational pull of the earth and never comes back to the earth is called the escape velocity.

Formula \(V_{esc}=\sqrt { \tfrac{2GM}{R}}\)

The force that acts on an object to keep it moving along the circular path with constant speed is called the centripetal force.

Its formula is \(F_c=\tfrac{mv^2}{r}\)

Answer::

Kepler's first law:The orbit of a planet is an ellipse with the Sun at one of the foci.Kepler's Second law:The line joining the planet and the Sun sweeps equal areas in equal intervals of timeKepler's Third law :The square of its period of revolution around the Sun is directly proportional to the cube of the mean distance of a planet from the Sun.d. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.

Let us consider an object which is projected vertically upwards with initial velocity u which reaches a maximum height h.

Acceleration due to gravity=-g

Equation of Motion for a body projected thrown upwards :

**V=u-gt----------(1)
h=ut-1/2gt² -----------(2)
v²-u²=-2gh --------(3)**

Equations of motion for freely falling body :

for free fall :

Initial velocity=u=0

V=gt----------(4)

h=\(1\over2\)gt²-------(5)

v²=2gh------(6)

Time of Ascent is the time taken by body thrown up to reach maximum height h

At maximum height , V=0

Equation (1) turns to u=gt1

\(t_1=\tfrac{u}{g} \) ----------(7)

Maximm height h=\(u²\over2g\) ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fall

so equation (5) h=\(1\over2\)gt₂²

t₂²=\(2h\over g\)

\(t₂=\sqrt{\tfrac{2h}{g}}\)

but from equation (9)

t₂=\( \sqrt{\tfrac{2xu²}{2g²}}\)

t₂=u/g -----------------equation (10)

The time ascent is equal to time of descent in case of bodies moving under gravity

Thus, the weight of the object or the pull of the floor on the object is W=mg

Now, if g becomes twice, then the weight of the object or the pull of the floor on the object also becomes twice

i.e.

it will become two times more difficult to pull it along the floor.

Acceleration due to the earth's gravity is zero at the centre of the Earth because at that point the mass of the earth is equally distributed in all directions, so pulling equally in all directions for a net zero pull. As the distance from the centre decreases, the acceleration due to gravity also decreases.

Hence , the value of g is zero at centre .

Hence , the value of g is zero at centre .

Thus, when the period of revolution of the planet at a distance R from a star is T, then from kepler's third law of planetry motion, we have

\(T^2\propto R^3 \space\space\space\space.........(1)\)

now , when the distance of the planet from the star is 2R, then its period of revolution becomes

\(T_1^2\propto(2R)^3\)

\(T_1^2\propto8R^3 \) ........(2)

Dividing equation 2 by 1

\(\tfrac{T_1^2}{T^2}=\tfrac{8R^3}{R^3}\)

\(T_1^2= 8 T^2\)

\(\therefore T_1=\sqrt8T\)

Answer::

**Given :** Time= 5s , Height=5m

**To find :** Gravitational acceleration

**Formula:** \(S=ut+\tfrac{1}{2}gt^2\)

**steps:**

by formula

\(5=0\times t+\tfrac{1}{2}g(5)^2\)

\(5=\tfrac{1}{2}g\times25\)

\(\therefore g=\tfrac{2}{5}\)

\(\therefore g=0.4 m/s^2\)