Gravitation – Test Papers

 CBSE Test Paper 01

Chapter 8 Gravitation

1. The direction of the universal gravitational force between particles of masses m1${\mathrm{m}}_1$ and m2${\mathrm{m}}_2$ is: 1
   1. towards m1${\mathrm{m}}_1$
   2. towards m2${\text{m}}_{\text{2}}$ on m1${\text{m}}_{\text{1}}$ and towards m1${\mathrm{m}}_1$ on m2${\mathrm{m}}_2$.
   3. towards the center of the earth
   4. towards m2${\mathrm{m}}_2$
2. The space shuttle releases a 470-kg communications satellite while in an orbit that is 280 km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit, which is an orbit in which the satellite stays directly over a single location on the Earth. How much energy did the engine have to provide? 1
   1. 1.09 ×$\times$ 1010$10^{10}$ J
   2. 1.29 ×$\times$ 1010$10^{10}$ J
   3. 1.39 ×$\times$ 1010$10^{10}$ J
   4. 1.19 ×$\times$ 1010$10^{10}$J
3. Time period of an earth satellite very close to the surface of earth is given by 1
   1. T0=π2REg−−−√$T_0=\pi \sqrt{\frac{2R_E}{g}}$
   2. T0=πRE2g−−−√$T_0=\pi \sqrt{\frac{R_E}{2g}}$
   3. T0=2πREg−−−√$T_0=2\pi \sqrt{\frac{R_E}{g}}$
   4. T0=πREg−−−√$T_0=\pi \sqrt{\frac{R_E}{g}}$
4. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? 1
   1. 18 N
   2. 128 N
   3. 28 N
   4. 180 N
5. Two stars each of one solar mass (= 2 ×$\times$ 1030$10^{30}$ kg) are approaching each other for a head on collision. When they are a distance 109$10^9$ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. assume the stars to remain undistorted until they collide. (Use the known value of G) 1
   1. 2.6 ×$\times$ 106$10^6$ m/s
   2. 1.6 ×$\times$ 106$10^6$ m/s
   3. 2.2 ×$\times$ 106$10^6$ m/s
   4. 2.8 ×$\times$ 106$10^6$ m/s

6. Do the friction of force and other contact forces arise due to gravitational attraction? If not, then what is the origin of these forces? 1

7. Define the effect of the shape of the earth on the value of g. 1

8. A thief with a box in his hand jumps from the top of a building. What will be the load experienced by him during the state of free fall? 1

9. If an object at the altitude of the space shuttle’s orbit, about 400 km about the earth’s surface, then find out the free fall acceleration of that object. 2

10. Derive an expression for work done against gravity. 2

11. State three essential requisites of geostationary satellite. 2

12. Find the distance of a point from the earth’s centre where the resultant gravitational field due to the earth and the moon is zero. The mass of the earth is 6.0 ×$\times$ 1024 kg and that of the moon is 7.4 ×$\times$ 1022 kg. The distance between the earth and the moon is 4.0 ×$\times$ 106 km. 3

13. Two uniform solid spheres of radii R and 2R are at rest with their surfaces just touching. Find the force of gravitational attraction between them if density of spheres be P? 3

14. Obtain an expression for escape velocity from energy considerations. 3

15. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2×1030$2\times {10}^{30}$kg; mass of mars = 6.4×1023$6.4\times {10}^{23}$ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28×108kg$2.28\times {10}^8\mathrm{k}\mathrm{g}$; G=6.67×10−11m2kg−2$G=6.67\times {10}^{-11}{\mathrm{m}}^2{\mathrm{k}\mathrm{g}}^{-2}$. 5

CBSE Test Paper 01
Chapter 8 Gravitation

Answer

1. 2. towards m2${\text{m}}_{\text{2}}$ on m1${\text{m}}_{\text{1}}$ and towards m1${\mathrm{m}}_1$ on m2${\mathrm{m}}_2$.
      Explanation: since gravitational force is attractive in nature, so if there are two particles of m1 & m2 .
      So One Force will be on m1 which will be directed towards m 2. i.e F→12=Gm1m2r212${\overrightarrow{F}}_{12}=G\frac{m_1{m}_2}{{r_{21}}^2}$
      And other force wiil be on m2 which will be directed towards m 1 i.e F→21=−Gm1m2r122${\overrightarrow{F}}_{21}=-G\frac{m_1{m}_2}{{r_{12}}^2}$
      Clearly, It can be seen that F→12=−F→21${\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21}$ Because thsese forces are attracted to each other and direction of each force is opposite to other one.

2. 4. 1.19 ×$\times$ 1010$10^{10}$J
      Explanation: Period of rocket in geosynchronous orbit is same as that of the earth:
      That is T = 1day = 24hours = 24 ×$\times$ 60 ×$\times$ 60 sec = 8.64 ×$\times$ 104 s
      From Keplers 3rd law
      T2 = KEr3GS
      Where KE = 4π2GME$\frac{4{\pi }^2}{G{M}_E}$ = 9.89 ×$\times$ 10-14s2/m3
      Therefore the geosynchronus radius is
      rGS=T2KE−−−√3=(8.64×104)29.89×10−14−−−−−−−−√3=4.23×107m$r_{GS}=\sqrt[3]{\frac{T^2}{K_E}}=\sqrt[3]{\frac{{\left(8.64\times {10}^4\right)}^2}{9.89\times {10}^{-14}}}=4.23\times {10}^{7}m$
      Because the initial position before the boost is 280 km = 2.8 ×$\times$ 105 m
      and the radius of the Earth is 6,370 km = 6.37 ×$\times$ 106 m
      Therefore ri = RE + 2.80 ×$\times$ 105 m = 6.65 ×$\times$ 106 m
      The total energy needed to boost the satellite at the geosynchronus radius is the difference of total energy before and after the boost
      E=−GMEms2(1rGS−1ri)$E=-\frac{G{M}_E{m}_s}{2}\left(\frac{1}{r_{GS}}-\frac{1}{r_i}\right)$
      =−6.67×10−11×5.98×1024×4702$=-\frac{6.67\times {10}^{-11}\times 5.98\times {10}^{24}\times 470}{2}$(14.23×107−16.65×106)$\left(\frac{1}{4.23\times {10}^7}-\frac{1}{6.65\times {10}^6}\right)$= 1.19 ×$\times$ 1010J
3. 3. T0=2πREg−−−√$T_0=2\pi \sqrt{\frac{R_E}{g}}$
      Explanation: It is the time taken by satellite to go once around the earth
      ∴T = Circumference of the orbitorbitalvelocity$\therefore \text{T = }\frac{\text{Circumference of the orbit}}{\text{orbitalvelocity}}$
      ⇒T=2πrν=2πrGM−−−√$\Rightarrow T=\frac{2\pi r}{\nu }=2\pi \sqrt{\frac{r}{GM}}$ [As ν = GMr −−−−√]$\text{[As }\nu \text{ = }\sqrt{\frac{GM}{r}}]$
      Since Satellite is very close to the surface of earth, So Here, we can take r=R(Radius of earth)
      ⇒T=2πR3GM−−−√$\Rightarrow T=2\pi \sqrt{\frac{R^3}{GM}}$
      As [GM=gR2]$[GM=g{R}^2]$
      ∴⇒T=2πR3gR2−−−√$\therefore \Rightarrow T=2\pi \sqrt{\frac{R^3}{g{R}^2}}$
      ⇒T=2πRg−−√$\Rightarrow T=2\pi \sqrt{\frac{R}{g}}$
4. 3. 28 N
      Explanation: gh=g(1+hR)−2$g_h=g{\left(1+\frac{h}{R}\right)}^{-2}$
      Given, h = R2$\frac{R}{2}$
      ∴gh=g(1+R2R)−2$\therefore g_h=g{\left(1+\frac{\frac{\text{R}}{2}}{R}\right)}^{-2}$
      ⇒ghg=1(1+12)2=1(32)2=(23)2=49$\Rightarrow \frac{g_h}{g}=\frac{1}{{\left(1+\frac{1}{2}\right)}^2}=\frac{1}{{\left(\frac{3}{2}\right)}^2}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$
      Let m = mass of the body
      If W and Wh be its weight at earth’s surface and at a height h above earth’s surface, then
      W = mg
      and Wh=mgh=m×49g=49mg$\text{and }{\text{W}}_h=m{g}_h=m\times \frac{4}{9}g=\frac{4}{9}mg$
      ⇒Wh=mgh=49×63=28N$\Rightarrow {\text{W}}_h=m{g}_h=\frac{4}{9}\times 63=28N$
5. 1. 2.6 ×106$\times {10}^6$ m/s
      Explanation: Let v⃗ 1${\overrightarrow{v}}_1$ and v⃗ 2${\overrightarrow{v}}_2$ be the velocities of two stars when they collide. 
      According to the law of conservation of momentum,
      Mv⃗ 1+Mv⃗ 2=0$M{\overrightarrow{v}}_1+M{\overrightarrow{v}}_2=0$
      ⇒v⃗ 1=−v⃗ 2$\Rightarrow {\overrightarrow{v}}_1=-{\overrightarrow{v}}_2$
      ⇒v1=v2=v$\Rightarrow v_1=v_2=v$
      According to law of conservation of energy,
      2[12mv2]=GM2[1r2−1r1]$\text{2}\left[\frac{1}{2}m{v}^2\right]=G{M}^2\left[\frac{1}{r_2}-\frac{1}{r_1}\right]$
      We Have
      Mass of the star, M = 2