Answer :
Let's solve the given questions one by one in a detailed manner:
26. An object having a mass of 24 kg on Earth is taken to a planet where the acceleration due to gravity is half of its value on Earth. The mass of the object on the planet is:
- The mass of an object does not change with the location. It remains the same regardless of whether it is on Earth or another planet.
- Therefore, the mass of the object on the planet is still [tex]\( \boxed{24 \text{ kg}} \)[/tex].
27. An object of mass 40 kg will experience a gravitational force of 68 N on a planet. What is the intensity of the gravitational field on this planet?
- The force experienced by an object in a gravitational field is given by [tex]\( F = mg \)[/tex], where [tex]\( F \)[/tex] is the force, [tex]\( m \)[/tex] is the mass, and [tex]\( g \)[/tex] is the gravitational field intensity.
- Given: [tex]\( F = 68 \text{ N} \)[/tex], [tex]\( m = 40 \text{ kg} \)[/tex]
- Solving for [tex]\( g \)[/tex]:
[tex]\[ g = \frac{F}{m} = \frac{68 \text{ N}}{40 \text{ kg}} = 1.7 \text{ N/kg} \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.7 \text{ N/kg}} \)[/tex].
28. The mass of a certain planet [tex]\( P \)[/tex] is one-hundredth that of Earth while its radius is one-quarter of Earth's radius. If the acceleration due to gravity on Earth is [tex]\( 10 \text{ m/s}^2 \)[/tex], what is its value on [tex]\( P \)[/tex]?
- Let [tex]\( M_e \)[/tex] and [tex]\( R_e \)[/tex] be the mass and radius of Earth, respectively.
- Given: [tex]\( M_p = \frac{M_e}{100} \)[/tex] and [tex]\( R_p = \frac{R_e}{4} \)[/tex]
- The acceleration due to gravity [tex]\( g \)[/tex] is given by [tex]\( g = \frac{GM}{R^2} \)[/tex].
- For planet [tex]\( P \)[/tex], the acceleration due to gravity [tex]\( g_p \)[/tex] can be found using:
[tex]\[ g_p = \frac{G \cdot M_p}{R_p^2} = \frac{G \cdot \frac{M_e}{100}}{\left(\frac{R_e}{4}\right)^2} = \frac{G \cdot \frac{M_e}{100}}{\frac{R_e^2}{16}} = \frac{G \cdot M_e \cdot 16}{100 \cdot R_e^2} = \frac{16 \cdot GM_e}{100 \cdot R_e^2} \][/tex]
- Since [tex]\( g_e = \frac{GM_e}{R_e^2} = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ g_p = \frac{16}{100} \cdot g_e = 0.16 \cdot 10 \text{ m/s}^2 = 1.6 \text{ m/s}^2 \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.6 \text{ m/s}^2} \)[/tex].
29. The escape velocity of a body on the surface of a planet of radius [tex]\( R \)[/tex] and mass [tex]\( M \)[/tex] is [ [tex]\( G \)[/tex] = universal gravitational constant ]:
- The formula for escape velocity [tex]\( v \)[/tex] is derived from the equation:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{\sqrt{\frac{2GM}{R}}} \)[/tex].
30. The radius of the Earth is [tex]\( 6.4 \times 10^6 \)[/tex] m and the acceleration due to gravity is [tex]\( 10.0 \text{ m/s}^2 \)[/tex]. The escape velocity of a rocket launched from the Earth's surface is:
- Using the escape velocity formula:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- As per the given data:
[tex]\[ v = 11313.70849898476 \text{ m/s} = 11.313708499 \text{ km/s} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{11.3 \text{ km/s}} \)[/tex].
26. An object having a mass of 24 kg on Earth is taken to a planet where the acceleration due to gravity is half of its value on Earth. The mass of the object on the planet is:
- The mass of an object does not change with the location. It remains the same regardless of whether it is on Earth or another planet.
- Therefore, the mass of the object on the planet is still [tex]\( \boxed{24 \text{ kg}} \)[/tex].
27. An object of mass 40 kg will experience a gravitational force of 68 N on a planet. What is the intensity of the gravitational field on this planet?
- The force experienced by an object in a gravitational field is given by [tex]\( F = mg \)[/tex], where [tex]\( F \)[/tex] is the force, [tex]\( m \)[/tex] is the mass, and [tex]\( g \)[/tex] is the gravitational field intensity.
- Given: [tex]\( F = 68 \text{ N} \)[/tex], [tex]\( m = 40 \text{ kg} \)[/tex]
- Solving for [tex]\( g \)[/tex]:
[tex]\[ g = \frac{F}{m} = \frac{68 \text{ N}}{40 \text{ kg}} = 1.7 \text{ N/kg} \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.7 \text{ N/kg}} \)[/tex].
28. The mass of a certain planet [tex]\( P \)[/tex] is one-hundredth that of Earth while its radius is one-quarter of Earth's radius. If the acceleration due to gravity on Earth is [tex]\( 10 \text{ m/s}^2 \)[/tex], what is its value on [tex]\( P \)[/tex]?
- Let [tex]\( M_e \)[/tex] and [tex]\( R_e \)[/tex] be the mass and radius of Earth, respectively.
- Given: [tex]\( M_p = \frac{M_e}{100} \)[/tex] and [tex]\( R_p = \frac{R_e}{4} \)[/tex]
- The acceleration due to gravity [tex]\( g \)[/tex] is given by [tex]\( g = \frac{GM}{R^2} \)[/tex].
- For planet [tex]\( P \)[/tex], the acceleration due to gravity [tex]\( g_p \)[/tex] can be found using:
[tex]\[ g_p = \frac{G \cdot M_p}{R_p^2} = \frac{G \cdot \frac{M_e}{100}}{\left(\frac{R_e}{4}\right)^2} = \frac{G \cdot \frac{M_e}{100}}{\frac{R_e^2}{16}} = \frac{G \cdot M_e \cdot 16}{100 \cdot R_e^2} = \frac{16 \cdot GM_e}{100 \cdot R_e^2} \][/tex]
- Since [tex]\( g_e = \frac{GM_e}{R_e^2} = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ g_p = \frac{16}{100} \cdot g_e = 0.16 \cdot 10 \text{ m/s}^2 = 1.6 \text{ m/s}^2 \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.6 \text{ m/s}^2} \)[/tex].
29. The escape velocity of a body on the surface of a planet of radius [tex]\( R \)[/tex] and mass [tex]\( M \)[/tex] is [ [tex]\( G \)[/tex] = universal gravitational constant ]:
- The formula for escape velocity [tex]\( v \)[/tex] is derived from the equation:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{\sqrt{\frac{2GM}{R}}} \)[/tex].
30. The radius of the Earth is [tex]\( 6.4 \times 10^6 \)[/tex] m and the acceleration due to gravity is [tex]\( 10.0 \text{ m/s}^2 \)[/tex]. The escape velocity of a rocket launched from the Earth's surface is:
- Using the escape velocity formula:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- As per the given data:
[tex]\[ v = 11313.70849898476 \text{ m/s} = 11.313708499 \text{ km/s} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{11.3 \text{ km/s}} \)[/tex].
