Q1. A _______ is necessary to change the speed as well as the direction of motion of an object.
A force is necessary to change the speed as well as the direction of motion of an object.
Updated on: 2026-03-31 | Author: Rahul Patil
A force is necessary to change the speed as well as the direction of motion of an object.
Let v be the velocity of the ball on reaching the ground.
Thus, from first equation of motion, we have
Let h be the height of the table. Thus, from the second equation of motion, we have
`"S" = "ut" + 1/2 "gt"^2`
`⇒ "h" = 0 + 1/2 xx 10xx1^2`
A person weighs 60 N on earth. His weight on the moon will be 10 N.
To find:
Formulae:
a = –g = –10 m/s2
Time for upward journey of the ball will be the same as time for downward journey i.e., 0.9 s.
To find: Gravitational acceleration (g)
5 = `0 xx 5 + 1/2 "g"(5)^2`
The gravitational acceleration of the planet is 0.4 m/s2.
From newton’s third equation of the motion;
Where;
According to our question;
The figure below illustrates the situation given in the question
Putting the above values we get
From the Newton’s first law of motion;
Where symbols have there usual meanings as above;
Putting the values we get;
t = `100/10` = 10s
Now we know that time required by an object to go up is same as time required to come down.
As we go above the earth's surface, value of g increases- Wrong
| Object | On Earth | On moon |
| Mass | X | X |
| Weight | 6Y | Y |
The acceleration produced in a body under the influence of the force of gravity alone is called acceleration due to gravity.
| Sr.no. | Universal gravitational constant | Gravitational acceleration of earth |
| 1. | The gravitational force acting between unit masses kept at a unit distance away from each other equals gravitational constant (G). | The acceleration produced in a body under the influence of the force of gravity alone is called gravitational acceleration of earth or acceleration due to gravity (g). |
| 2. | Gravitational constant is a scalar quantity. | Acceleration due to gravity is a vector quantity. |
| 3. | The value of a gravitational constant is a constant. | The value of acceleration due to gravity varies with height, depth and shape of the earth. |
| 4. | The value of G = 6.67 × 10-11 Nm2/kg2. | The value of g = 9.8 m/s2 on earth’s surface. |
| 5. | Gravitational constant is never zero anywhere. | Acceleration due to gravity is zero at the centre of the earth. |
| 6. | The S.I. unit of gravitational constant is Nm2/kg2 | The S.I. unit of acceleration due to gravity is m/s2. |
Free fall of an object does not depend on the mass of the object- Right
Ratio of mass of the earth (ME) to mass of the moon (MM)
`= "M"_"E"/"M"_"M"` = 81
Ratio of radius of the earth (RE) to radius of moon (RM)
`= "R"_"E"/"R"_"M"` = 3.7
To find: Weight on moon (Wmoon)
Calculation: Weight on the Earth,
mg = `"mGM"_"E"/"R"_"E"^2`
= 750
Weight on Moon,
`"W"_"moon" = "mGM"_"M"/"R"_"M"^2` ....(ii)
Substituting equation (i) in equation (ii),
`"W"_"moon" = (750 "R"_"E"^2)/(("GM"_"E")) xx "GM"_"M"/"R"_"M"^2`
`= 750 "R"_"E"^2/"R"_"M"^2 xx "M"_"M"/"M"_"E"`
`"W"_"moon" = 750 xx (3.7)^2 xx 1/81`
= 126.8 N
The weight of the person on moon will be 126.8 N.
If the distance between two masses is doubled, the gravitational force between them becomes less than the previous force- Right
To find: Gravitational force (F)
F = `(6.67 xx 10^-11 xx 75 xx 80)/1^2`
The gravitational force between Mahendra and Virat is 4.002 × 10-7 N.
This statement is Right.
The orbit of a planet is an ellipse with the Sun at one of the foci.
The closer a planet is to the Sun, the greater will be its velocity.
Area ASB, CSD and ESF are equal when the planet covers distance AB, CD, EF in the same time.
The stone held in the hand is stable because on it two balanced forces are exerted.
At the time of free fall on the earth, in addition to gravitational force, the object experiences force of friction due to air. Thus, the free fall cannot happen on the earth.
The velocity of the object increases due to acceleration due to the gravity of the earth.
During free fall only gravitational force is exerted on the object.
Object experiences free fall when the only force acting on the object is a gravitational force. During free fall in air, the frictional force due to air opposes the motion of the object and a buoyant force also acts on the object. Thus, true free fall is possible only in a vacuum.
Statement: Gravitational force between two bodies in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The gravitational force is much weaker than other forces in nature.
The CGS unit of G is dyne.cm2/g2 - Right
The free fall of an object is possible only in vacuum.
The gravitational force between two bodies is directly proportional to the product of the masses of those bodies and is inversely proportional to the square of the distance between them.
To find: Mass (mm), weight (Wm) on moon
Calculation: The mass of the object is independent of gravity and remains unchanged i.e., 5 kg.
On moon, the mass of the object is 5 kg and the weight is 8.17 N.
To find: Escape velocity (ve)X
Formulae:
`"V"_"esc" = sqrt((2"GM"_"e")/"R"_"e")`
`("V"_"esc")_"X" = sqrt((2"GM"_"X")/"R"_"X")`
`("V"_"esc")_"X"/"V"_"esc" = sqrt(("M"_"X" xx "R"_"e")/("M"_"e" xx "R"_"X"))`
`= sqrt(4 xx 1/2)`
= 1.414
= 11.2 × 103 × 1.414
= 15.84 × 103 m/s
The escape velocity of a body from the planet ‘X’ is 15.84 × 103 m/s.
F = `("G""m"_1"m"_2)/"r"^2`
where, Me and Ms are the masses of the Earth and the Sun, respectively. Using all the given values, we have
\[ \Rightarrow \text{m}_2 = \frac{(3 . 5 \times {10}^{22} ) \times (1 . 5 \times {10}^{11} )^2}{(6 .
7 \times {10}^{- 11} ) \times (6 \times {10}^{24} )}\]
`= (7.88 xx 10^44)/(40.2 xx 10^13)`
= 1.96 × 1030 kg
The mass of the Sun is 1.96 × 1030 kg.
the mass of the moon,
the distance between the earth and the moon,
= 3.84 × 105 × 1000 m
= 3.84 × 108 m
the force exerted by the earth on the moon is
F = `("G" "M" xx "m")/"d"^2`
= `(6.7 xx 10^-11 "Nm"^2"kg"^-2 xx 6 xx 10^24 "kg" xx 7.4 xx 10^22 "kg")/(3.84 xx 10^8 "m")^2`
= `((6.7 xx 10^-11) xx (6 xx 10^24) xx (7.4 xx 10^22))/(3.84 xx 10^8)^2`
= `(6.7 xx 6 xx 7.4)/(3.84 xx 3.84) xx 10^19`
= 2 × 1020 N.
Thus the force exerted by the earth on the moon is 2 × 1020 N.
The orbit of a planet revolving around a star is elliptical.
The acceleration due to gravity of a planet is given as
\[\text{g} = \frac{\text{GM}}{\text{r}^2}\]
For planet A:
\[\text{g}_\text{A} = \frac{\text{GM}_A}{\text{r}_\text{A}^2}\]
For planet B:
\[\text{g}_{B} = \frac{\text{GM}_\text{B}}{\text{r}_\text{B}^2}\]
Now,
\[\text{g}_\text{B} = \frac{1}{2} \text{g}_\text{A}\] ...(Given) or,
\[\frac{\text{GM}_\text{B}}{\text{r}_\text{B}^2} = \frac{\text{G M}_\text{A}}{2 \text{r}_\text{A}^2}\]
\[\Rightarrow \text{M}_\text{B} = \frac{\text{M}_\text{A} \text{r}_\text{B}^2}{2 \text{r}_\text{A}^2}\]
Given:
\[\text{r}_\text{A} = \frac{1}{2} \text{r}_\text{B}\]
\[\Rightarrow \text{M}_\text{B} = \frac{\text{M}_\text{A} \text{r}_\text{B}^2}{2(\frac{1}{2} \text{r}_\text{B})^2} = 2 \text{M}_\text{A}\]
Thus, the mass of planet B should be twice that of planet A.
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.
The value of g is highest at the equator- Wrong
This statement is Wrong.
Explanation:
The value of a gravitational constant (G) is always the same irrespective of place. However, according to Newton's law of universal gravitation, the acceleration of gravity (g) varies with mass and the distance from its centre.
The value of gravitational acceleration at the centre of earth is zero- Right
The value of gravitational acceleration (g) is 9.78 m/s2 at the equator.
The value of gravitational acceleration (g) is highest at the poles.
The value of universal gravitational constant (G) in the SI unit is 6.673 × 10-11 Nm2/kg2.
Explanation:
This constant appears in Newton's law of gravitation and determines the strength of the gravitational force between two masses.
The weight of any object on the moon is nearly 1/6 of the weight of the earth.
The minimum velocity with which a body should be projected from the surface of a planet or moon, so that it escapes from the gravitational influence of the planet or moon is called as escape velocity.
The escape velocity is different for different planets.
The motion of any object under the influence of the force of gravity alone is called free fall.
In free fall, the initial velocity of the object is zero and goes on increasing due to the acceleration due to gravity of the earth.
During free fall, the frictional force due to air opposes the motion of the object and a buoyant force also acts on the object.
Thus, true free fall is possible only in a vacuum.
Acceleration due to gravity value on the surface of earth is 9.8 m/s2.
The shape of the earth is not perfectly spherical. It is slightly flattened at the poles and bulged at the equator.
As a result, the radius of the earth at the poles is less than that at the equator.
The acceleration due to gravity (g) on earth’s surface is given as, g = `"GM"/"R"^2`. The value of g depends on the mass M of the earth and the radius R of the earth.
As we go inside the earth, our distance from the centre of the earth decreases and no longer remains equal to the radius of the earth (R).
Along-with the distance, the part of the earth which contributes towards the gravitational force felt also decreases, decreasing the value of (M).
Due to combined result of change in R and M, value of g becomes zero at the centre of the earth.
As the height of an object from the surface of the earth increases, the distance between the object and the centre of the earth (r) increases.
As a result, the value of gravitational acceleration (g) decreases as we go above the earth's surface.
The acceleration due to gravity (g) on earth’s surface is given as, g = `"GM"/"R"^2`. The value of g depends on the mass M of the earth and the radius R of the earth.
As we go inside the earth, our distance from the centre of the earth decreases and no longer remains equal to the radius of the earth (R).
Along-with the distance, the part of the earth which contributes towards the gravitational force felt also decreases, decreasing the value of (M).
Due to combined result of a change in R and M, value of g decreases as we go deep inside the earth.
For motion of an object thrown upwards, magnitude of g is the same throughout the motion but the velocity of the object decreases with the increase in height.
For object thrown upwards, the direction of acceleration due to gravity is opposite to the velocity of the object.
Also, acceleration due to gravity on the planet is given by gp = `"GM"_"P"/(("R"_"P")^2)`
Since MP and RP are different on different planets, the weight of an object will be different on different planets even though its mass remains constant.
But, the buoyant and frictional forces on the stone are much less than the weight of the stone and does not affect the speed of the stone much.
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 time.
Kepler’s third law: The square of the orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.