138676
A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$. The time period of second satellite is longer than the first one (approximately) by
1 $1.5 \%$
2 $0.5 \%$
3 $3 \%$
4 $1 \%$
5 $2 \%$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=1.01 \mathrm{R}$ We know that, From Kepler's law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1.01 \mathrm{R}}{\mathrm{R}}\right)^{3}=(1+0.01)^{3}$ By Binomial Expression $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=(1+0.01)^{3 / 2}=\left(1+\frac{3}{2} \times 0.001\right)$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1.015$ $\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \times 100=(1.015-1) \times 100$ $=1.5 \%$
AIIMS 2002
Gravitation
138677
The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$, then the escape velocity from the planet is:
1 $\sqrt{3} \mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $\mathrm{V}$
4 $\sqrt{5} \mathrm{v}$
5 $\sqrt{12} \mathrm{v}$
Explanation:
A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively. Then $M_{p}=6 M_{e}, R_{p}=2 R_{e}$ We know that, escape velocity of earth - $\because \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity from the planet, $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}_{\mathrm{e}}}{2 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{V}_{\mathrm{P}}=\sqrt{\frac{6 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $=\sqrt{\frac{2 \times 3 \times \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\left\{\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}\right\}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}$
Kerala CEE 2006
Gravitation
138678
The escape velocity for the earth is $v_{e^{*}}$. The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is
1 $\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}$
2 $\sqrt{2} \mathrm{v}_{\mathrm{e}}$
3 $2 \mathrm{v}_{\mathrm{e}}$
4 $\frac{\mathrm{v}_{\mathrm{e}}}{2}$
Explanation:
B Given, $\text { Radius of planet }\left(R_p\right)=\frac{R_e}{4}$ $\text { Mass of planet }\left(M_p\right)=\frac{M_e}{2}$ Escape velocity of a body from the surface of the earth is given by $v_e=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Then escape velocity of a body from the surface of the planet is given by $v_p=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ \(=\sqrt{\frac{2 \mathrm{G} \times \mathrm{M}_{\mathrm{e}} \times 4}{2 \mathrm{R}_{\mathrm{e}}}}\) \(=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \times \sqrt{2}\) From equation (i) we get - \(=\sqrt{2} \mathrm{v}_{\mathrm{e}}\)
UPSEE - 2014
Gravitation
138679
The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be
1 $\sqrt{\mathrm{gR}}$
2 $\sqrt{0.5 \mathrm{gR}}$
3 $\sqrt{2 \mathrm{gR}}$
4 $\sqrt{3 g R}$
Explanation:
A According to law of universal gravity $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ And, $\mathrm{F}=\mathrm{mg}$ (According to Newton's second law of motion $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\mathrm{mg}$ $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$ We know that orbital velocity $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $\mathrm{R}$ of earth will be $\sqrt{\mathrm{gR}}$.
138676
A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$. The time period of second satellite is longer than the first one (approximately) by
1 $1.5 \%$
2 $0.5 \%$
3 $3 \%$
4 $1 \%$
5 $2 \%$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=1.01 \mathrm{R}$ We know that, From Kepler's law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1.01 \mathrm{R}}{\mathrm{R}}\right)^{3}=(1+0.01)^{3}$ By Binomial Expression $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=(1+0.01)^{3 / 2}=\left(1+\frac{3}{2} \times 0.001\right)$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1.015$ $\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \times 100=(1.015-1) \times 100$ $=1.5 \%$
AIIMS 2002
Gravitation
138677
The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$, then the escape velocity from the planet is:
1 $\sqrt{3} \mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $\mathrm{V}$
4 $\sqrt{5} \mathrm{v}$
5 $\sqrt{12} \mathrm{v}$
Explanation:
A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively. Then $M_{p}=6 M_{e}, R_{p}=2 R_{e}$ We know that, escape velocity of earth - $\because \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity from the planet, $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}_{\mathrm{e}}}{2 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{V}_{\mathrm{P}}=\sqrt{\frac{6 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $=\sqrt{\frac{2 \times 3 \times \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\left\{\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}\right\}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}$
Kerala CEE 2006
Gravitation
138678
The escape velocity for the earth is $v_{e^{*}}$. The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is
1 $\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}$
2 $\sqrt{2} \mathrm{v}_{\mathrm{e}}$
3 $2 \mathrm{v}_{\mathrm{e}}$
4 $\frac{\mathrm{v}_{\mathrm{e}}}{2}$
Explanation:
B Given, $\text { Radius of planet }\left(R_p\right)=\frac{R_e}{4}$ $\text { Mass of planet }\left(M_p\right)=\frac{M_e}{2}$ Escape velocity of a body from the surface of the earth is given by $v_e=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Then escape velocity of a body from the surface of the planet is given by $v_p=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ \(=\sqrt{\frac{2 \mathrm{G} \times \mathrm{M}_{\mathrm{e}} \times 4}{2 \mathrm{R}_{\mathrm{e}}}}\) \(=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \times \sqrt{2}\) From equation (i) we get - \(=\sqrt{2} \mathrm{v}_{\mathrm{e}}\)
UPSEE - 2014
Gravitation
138679
The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be
1 $\sqrt{\mathrm{gR}}$
2 $\sqrt{0.5 \mathrm{gR}}$
3 $\sqrt{2 \mathrm{gR}}$
4 $\sqrt{3 g R}$
Explanation:
A According to law of universal gravity $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ And, $\mathrm{F}=\mathrm{mg}$ (According to Newton's second law of motion $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\mathrm{mg}$ $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$ We know that orbital velocity $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $\mathrm{R}$ of earth will be $\sqrt{\mathrm{gR}}$.
138676
A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$. The time period of second satellite is longer than the first one (approximately) by
1 $1.5 \%$
2 $0.5 \%$
3 $3 \%$
4 $1 \%$
5 $2 \%$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=1.01 \mathrm{R}$ We know that, From Kepler's law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1.01 \mathrm{R}}{\mathrm{R}}\right)^{3}=(1+0.01)^{3}$ By Binomial Expression $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=(1+0.01)^{3 / 2}=\left(1+\frac{3}{2} \times 0.001\right)$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1.015$ $\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \times 100=(1.015-1) \times 100$ $=1.5 \%$
AIIMS 2002
Gravitation
138677
The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$, then the escape velocity from the planet is:
1 $\sqrt{3} \mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $\mathrm{V}$
4 $\sqrt{5} \mathrm{v}$
5 $\sqrt{12} \mathrm{v}$
Explanation:
A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively. Then $M_{p}=6 M_{e}, R_{p}=2 R_{e}$ We know that, escape velocity of earth - $\because \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity from the planet, $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}_{\mathrm{e}}}{2 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{V}_{\mathrm{P}}=\sqrt{\frac{6 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $=\sqrt{\frac{2 \times 3 \times \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\left\{\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}\right\}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}$
Kerala CEE 2006
Gravitation
138678
The escape velocity for the earth is $v_{e^{*}}$. The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is
1 $\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}$
2 $\sqrt{2} \mathrm{v}_{\mathrm{e}}$
3 $2 \mathrm{v}_{\mathrm{e}}$
4 $\frac{\mathrm{v}_{\mathrm{e}}}{2}$
Explanation:
B Given, $\text { Radius of planet }\left(R_p\right)=\frac{R_e}{4}$ $\text { Mass of planet }\left(M_p\right)=\frac{M_e}{2}$ Escape velocity of a body from the surface of the earth is given by $v_e=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Then escape velocity of a body from the surface of the planet is given by $v_p=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ \(=\sqrt{\frac{2 \mathrm{G} \times \mathrm{M}_{\mathrm{e}} \times 4}{2 \mathrm{R}_{\mathrm{e}}}}\) \(=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \times \sqrt{2}\) From equation (i) we get - \(=\sqrt{2} \mathrm{v}_{\mathrm{e}}\)
UPSEE - 2014
Gravitation
138679
The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be
1 $\sqrt{\mathrm{gR}}$
2 $\sqrt{0.5 \mathrm{gR}}$
3 $\sqrt{2 \mathrm{gR}}$
4 $\sqrt{3 g R}$
Explanation:
A According to law of universal gravity $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ And, $\mathrm{F}=\mathrm{mg}$ (According to Newton's second law of motion $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\mathrm{mg}$ $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$ We know that orbital velocity $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $\mathrm{R}$ of earth will be $\sqrt{\mathrm{gR}}$.
138676
A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$. The time period of second satellite is longer than the first one (approximately) by
1 $1.5 \%$
2 $0.5 \%$
3 $3 \%$
4 $1 \%$
5 $2 \%$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=1.01 \mathrm{R}$ We know that, From Kepler's law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1.01 \mathrm{R}}{\mathrm{R}}\right)^{3}=(1+0.01)^{3}$ By Binomial Expression $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=(1+0.01)^{3 / 2}=\left(1+\frac{3}{2} \times 0.001\right)$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1.015$ $\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1}} \times 100=(1.015-1) \times 100$ $=1.5 \%$
AIIMS 2002
Gravitation
138677
The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$, then the escape velocity from the planet is:
1 $\sqrt{3} \mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $\mathrm{V}$
4 $\sqrt{5} \mathrm{v}$
5 $\sqrt{12} \mathrm{v}$
Explanation:
A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively. Then $M_{p}=6 M_{e}, R_{p}=2 R_{e}$ We know that, escape velocity of earth - $\because \quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity from the planet, $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}_{\mathrm{e}}}{2 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{V}_{\mathrm{P}}=\sqrt{\frac{6 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $=\sqrt{\frac{2 \times 3 \times \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\left\{\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}\right\}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{3} \mathrm{v}$
Kerala CEE 2006
Gravitation
138678
The escape velocity for the earth is $v_{e^{*}}$. The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is
1 $\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}$
2 $\sqrt{2} \mathrm{v}_{\mathrm{e}}$
3 $2 \mathrm{v}_{\mathrm{e}}$
4 $\frac{\mathrm{v}_{\mathrm{e}}}{2}$
Explanation:
B Given, $\text { Radius of planet }\left(R_p\right)=\frac{R_e}{4}$ $\text { Mass of planet }\left(M_p\right)=\frac{M_e}{2}$ Escape velocity of a body from the surface of the earth is given by $v_e=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Then escape velocity of a body from the surface of the planet is given by $v_p=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ \(=\sqrt{\frac{2 \mathrm{G} \times \mathrm{M}_{\mathrm{e}} \times 4}{2 \mathrm{R}_{\mathrm{e}}}}\) \(=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \times \sqrt{2}\) From equation (i) we get - \(=\sqrt{2} \mathrm{v}_{\mathrm{e}}\)
UPSEE - 2014
Gravitation
138679
The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be
1 $\sqrt{\mathrm{gR}}$
2 $\sqrt{0.5 \mathrm{gR}}$
3 $\sqrt{2 \mathrm{gR}}$
4 $\sqrt{3 g R}$
Explanation:
A According to law of universal gravity $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ And, $\mathrm{F}=\mathrm{mg}$ (According to Newton's second law of motion $\frac{\mathrm{GMm}}{\mathrm{R}^{2}}=\mathrm{mg}$ $\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\mathrm{g}$ We know that orbital velocity $\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}=\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}$ $\mathrm{v}=\sqrt{\mathrm{gR}}$ Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $\mathrm{R}$ of earth will be $\sqrt{\mathrm{gR}}$.
