### Solution to problems – class 9 – Set 1 Q42, Q43, Q44, Q45, Q46

##### June 19, 2019

## Problem Statement

**42) A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? ( g = 9.8 m/ s^2).**

**43) A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone ?What is its mass ? (**

*g*= 10 m/s^2).**44)**

**(a) An object has mass of 20 kg on earth.**

**What will be its (**

*i*) mass, and (*ii*) weight, on the moon?(*g*on moon = 1.6 m/s^2).**(**10

*b*) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10^22 kg and a radius of 1.74 × 10^6 m (*G*= 6.7 ×^{-11}

**Nm^2/kg^2).**

**45) The mass of a planet is 6 × 10^24 kg and its diameter is 12.8 × 10^3 km. If the value of gravitational constant be 6.7 ×**10

^{-11}

**Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet.**

**46) The values of**

*g*at six distances A, B, C, D, E and F from the surface of the earth are found to be 3.08 m/s^2, 9.23 m/s^2, 0.57 m/s^2, 7.34 m/s^2, 0.30 m/s^2 and 1.49 m/s^2, respectively.**(**

*a*) Arrange these values of*g*according to the increasing distances from the surface of the earth (keeping the value of*g*nearest to the surface of the earth first)**(**

*b*) If the value of distance F be 10000 km from the surface of the earth(*hint: check if possible or not*), state whether this distance is deep inside the earth or high up in the sky.**Give reason for your answer of the place of observation**

## Solution

**Solution (Q42)**

Weight = W = 9.8 N, *g *= 9.8 m s^{-2}

Weight (W) = mass * acceleration due to gravity

W= mg

Mass of the body(m) = W / g = 9.8 /9.8 = 1 kg

Force = mass x acceleration

so, Acceleration = force /mass = 20 / 1 = 20 m/s^{2}

**Solution (Q43) **

Weight = gravitational force = 20 N

Mass = weight / g = 20/10 = 2 kg

**Solution (Q44) **

(a) i) Mass on the moon = 20 Kg

ii) Weight on the moon = m g_{moon}= 20 * 1.6 = 32 N

(b) say, g_{sat} is the acceleration due to gravity on the surface of the satellite.

Then, g_{sat} = GM/r^{2} = (6.7 × 10 ^{-11} x 7.4 × 10^{22}) / (1.74 × 10^{6})^{2} = 1.63 m/s^{2}

[We get the eqn. of g_{sat} in this way:

weight of an object of mass m on the satellite is mg_{sat }

The gravitational force on that mass m due the satellite of M is GMm/r^{2},

where r is the radius of the satellite.

weight of the object on satellite = gravitational force on the object due to the satellite

mg_{sat} = GMm/r^{2}

g_{sat} = GM/r^{2} ]

** Solution (Q45) **

**Diameter of the planet= 12 .8 × 10^{3 }km = 12.8 x 10^{6} m **

Radius = 6.4 x 10^{6} m

**g _{planet} = GM/r^{2} = (6.7× **

**10**x

^{-11}**6× 10**

^{24}) / (6.4 × 10^{6})^{2}**= 9.8 m/s ^{2}**

** Solution (Q46) **

**Solution(a) As we know, as height increases fromthe surface of earth the value of g decreases.**

**The values of g according to theincreasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first) are 9.23, 7.34, 3.08, 1.49,0.57,0.3 (all with units m/s^{2}) **

(b) The diameter of earth is 6400 km only. So this 10000 km distance can’t be

deep inside the earth. This must be high up in the sky.